Q.Verify that the function $y = e^{-3x}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

Showing the 12 most recent of 86 on this concept.

1. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If the differential equation having $y = Ae^x + B\sin x$ as its general solution is $f(x)\dfrac{d^2y}{dx^2} + g(x)\dfrac{dy}{dx} + h(x)y = 0$, then $f(x) + g(x) + h(x) =$ (A) $\cos x - \sin x$ (B) $4\sin x$ (C) $2\cos x$ (D) $0$

   ›Reveal solutionSolution

   Eliminating $A,B$ from $y=Ae^x+B\sin x$ gives $(\cos x-\sin x)y''+2\sin x\,y'-(\sin x+\cos x)y=0$, so $f+g+h=0$ — option (D).

   Concept. A two‑parameter family $y=Ae^{x}+B\sin x$ satisfies a second‑order ODE obtained by eliminating $A$ and $B$ from $y,\,y',\,y''$. The Wronskian‑style determinant of $\{y,e^{x},\sin x\}$ vanishing is exactly that eliminant.

   Solution.

   1. Differentiate:

   $y=Ae^{x}+B\sin x,\qquad y'=Ae^{x}+B\cos x,\qquad y''=Ae^{x}-B\sin x.$

   2. A non‑trivial $(A,B)$ exists iff

   $\begin{vmatrix} y & e^{x} & \sin x\\y' & e^{x} & \cos x\\y'' & e^{x} & -\sin x\end{vmatrix}=0.$

   3. Expand along the first column and divide by $e^{x}$:

   $(\cos x-\sin x)\,y'' + 2\sin x\,y' - (\sin x+\cos x)\,y = 0.$

   4. Read off the coefficients:

   $f(x)=\cos x-\sin x,\quad g(x)=2\sin x,\quad h(x)=-(\sin x+\cos x).$

   5. Add them:

   $f+g+h=(\cos x-\sin x)+2\sin x-(\sin x+\cos x)=0.$

   ✓Final answer

   $f(x)+g(x)+h(x)=0$ — option (D).

2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.The integrating factor of the linear differential equation in $x$ given by $\dfrac{dy}{dx} = \dfrac{1}{3x + y + 2}$ is (A) $e^{-3x}$ (B) $e^{-x}$ (C) $e^{-3y}$ (D) $e^{-y}$

   ›Reveal solutionSolution

   The given equation is not linear in $y$, but rewriting it as $\frac{dx}{dy} = 3x + y + 2$ makes it linear in $x$ with integrating factor $e^{-3y}$, so the correct option is (C).

   We are given

   $\frac{dy}{dx} = \frac{1}{3x + y + 2}.$

   At first glance, this looks like a first-order differential equation in $y$ as a function of $x$. But it is not linear in $y$ because the right-hand side is a rational function containing $y$ in the denominator. However, we can flip the relationship: treat $x$ as a function of $y$ instead.

   The key insight: if $\frac{dy}{dx}$ is given, then $\frac{dx}{dy} = 1 / \frac{dy}{dx}$ (provided the derivative is nonzero). This often turns a nonlinear equation in $y$ into a linear equation in $x$.

   ---

   1. Rewrite the equation in terms of $x(y)$ Since $\frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$, we have

   $\frac{dx}{dy} = 3x + y + 2.$

   This is now a linear first-order differential equation in $x$ with independent variable $y$.

   2. Identify the standard linear form The standard form for a linear ODE in $x$ is

   $\frac{dx}{dy} + P(y)\,x = Q(y).$

   Our equation is

   $\frac{dx}{dy} - 3x = y + 2.$

   So $P(y) = -3$ and $Q(y) = y + 2$.

   3. Recall the integrating factor formula For a linear ODE $\frac{dx}{dy} + P(y)x = Q(y)$, the integrating factor is

   $\mu(y) = e^{\int P(y)\,dy}.$

   Here $P(y) = -3$, so

   $\mu(y) = e^{\int (-3)\,dy} = e^{-3y}.$

   4. Interpret the result The integrating factor depends only on $y$, and it is $e^{-3y}$. Multiplying the equation by this factor makes the left side an exact derivative, allowing us to solve for $x(y)$.

   Thus, among the given options, the integrating factor is $e^{-3y}$.

   Watch out

   A common mistake is to try to force the equation into the form $\frac{dy}{dx} + P(x)y = Q(x)$ without checking if it is linear in $y$. Here it is not, so flipping variables is essential.

   Tip

   Whenever you see $\frac{dy}{dx} = \frac{1}{\text{linear in }x\text{ and }y}$, try writing $\frac{dx}{dy}$ instead — it often becomes linear immediately.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The general solution of the differential equation $(2xy + y^2)dy = (x^2 - y^2)dx$ is (A) $x^3 - 3x^2y - y^3 = c$ (B) $x^3 - 3x^2y + y^3 = c$ (C) $x^3 - 3xy^2 + y^3 = c$ (D) $x^3 - 3xy^2 - y^3 = c$

   ›Reveal solutionSolution

   This is a homogeneous differential equation. Substituting $y = vx$ reduces it to a separable form, and integrating gives the general solution $x^3 - 3xy^2 - y^3 = c$, which matches option (D).

   The given equation is $(2xy + y^2)dy = (x^2 - y^2)dx$. Notice that every term is of degree 2 — $2xy$, $y^2$, $x^2$, $y^2$ — so the equation is homogeneous. For a homogeneous equation, the standard trick is to set $y = vx$, which turns the equation into one where variables separate cleanly.

   1. Rewrite in standard form Bring the $dx$ term to the left:

   $(2xy + y^2)dy - (x^2 - y^2)dx = 0$

   Or equivalently,

   $\frac{dy}{dx} = \frac{x^2 - y^2}{2xy + y^2}$

   2. Substitute $y = vx$ Then $\frac{dy}{dx} = v + x\frac{dv}{dx}$. The right-hand side becomes:

   $\frac{x^2 - (vx)^2}{2x(vx) + (vx)^2} = \frac{x^2(1 - v^2)}{x^2(2v + v^2)} = \frac{1 - v^2}{2v + v^2}$

   So the equation is:

   $v + x\frac{dv}{dx} = \frac{1 - v^2}{2v + v^2}$

   3. Separate variables Subtract $v$ from both sides:

   $x\frac{dv}{dx} = \frac{1 - v^2}{2v + v^2} - v = \frac{1 - v^2 - v(2v + v^2)}{2v + v^2}$

   Simplify the numerator:

   $1 - v^2 - 2v^2 - v^3 = 1 - 3v^2 - v^3$

   So:

   $x\frac{dv}{dx} = \frac{1 - 3v^2 - v^3}{2v + v^2}$

   Now separate:

   $\frac{2v + v^2}{1 - 3v^2 - v^3} \, dv = \frac{dx}{x}$

   4. Integrate both sides The left-hand side is set up for a simple substitution. Let $u = 1 - 3v^2 - v^3$. Then $du = (-6v - 3v^2)dv = -3(2v + v^2)dv$. Notice that $2v + v^2$ appears in the numerator, so:

   $\frac{2v + v^2}{1 - 3v^2 - v^3} \, dv = -\frac{1}{3} \frac{du}{u}$

   Integrating:

   $\int -\frac{1}{3} \frac{du}{u} = \int \frac{dx}{x}$

   $-\frac{1}{3} \log|u| = \log|x| + C$

   Multiply by $-3$:

   $\log|u| = -3\log|x| - 3C$

   $\log|u| = \log|x^{-3}| + \log K \quad (\text{where } \log K = -3C)$

   So:

   $u = \frac{K}{x^3}$

   5. Back-substitute Recall $u = 1 - 3v^2 - v^3$ and $v = y/x$:

   $1 - 3\left(\frac{y}{x}\right)^2 - \left(\frac{y}{x}\right)^3 = \frac{K}{x^3}$

   Multiply through by $x^3$:

   $x^3 - 3xy^2 - y^3 = K$

   Renaming the constant $K$ as $c$, we get the general solution:

   $x^3 - 3xy^2 - y^3 = c$

   Watch out

   A common mistake is to misplace the signs when simplifying the numerator after subtracting $v$. Always combine terms carefully: $v = \frac{v(2v+v^2)}{2v+v^2}$, so the numerator becomes $1 - v^2 - 2v^2 - v^3 = 1 - 3v^2 - v^3$.

   Tip

   Spotting the derivative pattern $du = -3(2v+v^2)dv$ saves time — it means you don't need partial fractions or any heavy algebra.

   ✓Final answer

   The correct option is (D): $x^3 - 3xy^2 - y^3 = c$.

4. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.The general solution of $\dfrac{dy}{dx} + y f'(x) - f(x) f'(x) = 0, \; y \neq f(x)$ is (A) $y = f(x) + 1 + c e^{-f(x)}$ (B) $y = c e^{-f(x)}$ (C) $y = f(x) - 1 + c e^{-f(x)}$ (D) $y = f(x) + c e^{f(x)}$

   ›Reveal solutionSolution

   This is a first-order linear ODE disguised by the presence of $f(x)$ and $f'(x)$. By rewriting it in standard form and using an integrating factor $e^{f(x)}$, the general solution simplifies to $y = f(x) - 1 + c e^{-f(x)}$, which corresponds to option (C).

   We start with the given differential equation:

   $\frac{dy}{dx} + y f'(x) - f(x) f'(x) = 0, \quad y \neq f(x).$

   ---

   Concept & Intuition

   The equation looks messy because of the $f(x)$ and $f'(x)$ terms, but notice that $f'(x)$ appears as a coefficient of $y$ and also multiplied by $f(x)$. This suggests we can rearrange it into the standard linear form $\frac{dy}{dx} + P(x) y = Q(x)$, where $P(x)$ and $Q(x)$ are functions of $x$ only. Once in that form, the method of integrating factor works cleanly. The key trick: treat $f(x)$ as some known function (we don't need its explicit form), and $f'(x)$ as its derivative.

   ---

   Step-by-step solution

   1. Rewrite the equation in standard linear form Bring the term $- f(x) f'(x)$ to the right-hand side:

   $\frac{dy}{dx} + f'(x) \, y = f(x) f'(x).$

   This is now of the form $\frac{dy}{dx} + P(x) y = Q(x)$ with $P(x) = f'(x)$ and $Q(x) = f(x) f'(x)$.

   2. Find the integrating factor The integrating factor $\mu(x)$ is given by $e^{\int P(x) \, dx}$. Here:

   $\int P(x) \, dx = \int f'(x) \, dx = f(x) + C.$

   We only need one integrating factor, so take $\mu(x) = e^{f(x)}$.

   3. Multiply through by the integrating factor

   $e^{f(x)} \frac{dy}{dx} + e^{f(x)} f'(x) \, y = e^{f(x)} f(x) f'(x).$

   The left-hand side is the derivative of $y e^{f(x)}$ with respect to $x$ (by the product rule, since $\frac{d}{dx} e^{f(x)} = e^{f(x)} f'(x)$). So we have:

   $\frac{d}{dx} \left( y e^{f(x)} \right) = e^{f(x)} f(x) f'(x).$

   4. Integrate both sides

   $y e^{f(x)} = \int e^{f(x)} f(x) f'(x) \, dx.$

   Notice that the integrand is set up for a substitution: let $u = f(x)$, then $du = f'(x) \, dx$, so:

   $\int e^{u} u \, du.$

   This is a standard integral. Use integration by parts: let $w = u$, $dv = e^u du$, then $dw = du$, $v = e^u$. So:

   $\int u e^u \, du = u e^u - \int e^u \, du = u e^u - e^u + C = e^u (u - 1) + C.$

   Substituting back $u = f(x)$:

   $\int e^{f(x)} f(x) f'(x) \, dx = e^{f(x)} \big( f(x) - 1 \big) + C.$

   5. Solve for $y$ From step 4:

   $y e^{f(x)} = e^{f(x)} \big( f(x) - 1 \big) + C.$

   Divide both sides by $e^{f(x)}$ (which is never zero):

   $y = f(x) - 1 + C e^{-f(x)}.$

   Renaming the constant $C$ as $c$, we get the general solution:

   $y = f(x) - 1 + c e^{-f(x)}.$

   ---

   Watch out

   A common mistake is to forget the constant of integration or to misapply integration by parts. Also note the condition $y \neq f(x)$ is given to avoid the trivial case where the denominator in some step might vanish, but it doesn't affect the derivation here.

   Tip

   The structure $y = f(x) - 1 + c e^{-f(x)}$ shows that as $x$ varies, the term $c e^{-f(x)}$ decays or grows depending on $f(x)$, but the particular solution $f(x)-1$ is the "steady" part.

   ---

   ✓Final answer

   The correct option is (C).

   ANSWER: C

5. TG EAPCET 2021Set eng-2021-08-05-FN1 markMCQ
   Q.If $x = 2 + 2^{2/3} + 2^{1/3}$, then $x^3 - 6x^2 + 6x =$ (A) 3 (B) 2 (C) 1 (D) 0

   ›Reveal solutionSolution

   The key idea is to rewrite the given expression as a cubic equation by setting $t = 2^{1/3}$, then substituting $x = t^2 + t + 2$ and simplifying to find $x^3 - 6x^2 + 6x = 2$.

   We are given $x = 2 + 2^{2/3} + 2^{1/3}$. The expression $x^3 - 6x^2 + 6x$ looks like it might simplify to a constant — a common trick in such problems. The direct approach of cubing $x$ would be messy, so we look for a smarter algebraic structure.

   Notice that $2^{1/3}$ and $2^{2/3}$ are related: if we let $t = 2^{1/3}$, then $t^3 = 2$ and $t^2 = 2^{2/3}$. So $x$ becomes $x = t^2 + t + 2$. The problem now reduces to working with $t$, where $t^3 = 2$ is a clean relation.

   1. Set up the substitution.

      Let $t = 2^{1/3}$. Then $t^3 = 2$, and $x = t^2 + t + 2$.

   2. Express $x$ in terms of $t$ and find a relation.

      We have $x - 2 = t^2 + t$.

      Square both sides? That would give $t^4 + 2t^3 + t^2$, but $t^4 = t \cdot t^3 = 2t$, so it might work. But a more elegant path is to notice that $t$ satisfies $t^3 - 2 = 0$, and $x$ is a symmetric polynomial in $t$. We can compute $x^3 - 6x^2 + 6x$ directly by substituting $x = t^2 + t + 2$ and simplifying using $t^3 = 2$.

   3. Compute $x^2$ and $x^3$ in terms of $t$.

      First, $x^2 = (t^2 + t + 2)^2$

      $= t^4 + t^2 + 4 + 2t^3 + 4t^2 + 4t$ (expand carefully: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$)

      $= t^4 + 2t^3 + (t^2 + 4t^2) + 4t + 4$

      $= t^4 + 2t^3 + 5t^2 + 4t + 4$.

      Now $t^4 = t \cdot t^3 = 2t$, and $t^3 = 2$, so:

      $x^2 = 2t + 2(2) + 5t^2 + 4t + 4$

      $= 2t + 4 + 5t^2 + 4t + 4$

      $= 5t^2 + 6t + 8$.

      Next, $x^3 = x \cdot x^2 = (t^2 + t + 2)(5t^2 + 6t + 8)$.

      Multiply term by term:

      $t^2(5t^2 + 6t + 8) = 5t^4 + 6t^3 + 8t^2$

      $t(5t^2 + 6t + 8) = 5t^3 + 6t^2 + 8t$

      $2(5t^2 + 6t + 8) = 10t^2 + 12t + 16$

      Summing: $x^3 = 5t^4 + (6t^3 + 5t^3) + (8t^2 + 6t^2 + 10t^2) + (8t + 12t) + 16$

      $= 5t^4 + 11t^3 + 24t^2 + 20t + 16$.

      Replace $t^4 = 2t$ and $t^3 = 2$:

      $x^3 = 5(2t) + 11(2) + 24t^2 + 20t + 16$

      $= 10t + 22 + 24t^2 + 20t + 16$

      $= 24t^2 + 30t + 38$.

   4. Now form $x^3 - 6x^2 + 6x$.

      We have:

      $x^3 = 24t^2 + 30t + 38$

      $6x^2 = 6(5t^2 + 6t + 8) = 30t^2 + 36t + 48$

      $6x = 6(t^2 + t + 2) = 6t^2 + 6t + 12$

      So:

      $x^3 - 6x^2 + 6x = (24t^2 + 30t + 38) - (30t^2 + 36t + 48) + (6t^2 + 6t + 12)$

      Combine $t^2$ terms: $24 - 30 + 6 = 0$

      Combine $t$ terms: $30 - 36 + 6 = 0$

      Constant terms: $38 - 48 + 12 = 2$

      All $t$ terms cancel perfectly, leaving just $2$.

   Watch out

   A common mistake is to try cubing $x$ directly without substitution — that leads to a mess with $2^{2/3}$ and $2^{1/3}$ terms that are hard to simplify. The substitution $t = 2^{1/3}$ makes the algebra clean because $t^3 = 2$ is a simple number.

   Tip

   Notice that the coefficients $1, -6, 6$ in $x^3 - 6x^2 + 6x$ are suspiciously close to those in $(x-2)^3 = x^3 - 6x^2 + 12x - 8$. Indeed, $x^3 - 6x^2 + 6x = (x-2)^3 - 6x + 8 + 6x = (x-2)^3 + 8$? Let's check: $(x-2)^3 = x^3 - 6x^2 + 12x - 8$, so $x^3 - 6x^2 + 6x = (x-2)^3 - 6x + 8$. That doesn't simplify directly. But the substitution method is foolproof.

   ✓Final answer

   The value is $\boxed{2}$, which corresponds to option (B).

6. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.If $\cos x \dfrac{dy}{dx} = y \sin x - 1$, $x \neq (2n+1)\dfrac{\pi}{2}$, $n \in \mathbb{Z}$ is the differential equation corresponding to the curve $y = f(x)$ and $f(0) = 1$ then $f(x) =$ (A) $(1-x)\sec x$ (B) $(1-x)\cos x$ (C) $x + \cos x$ (D) $x + \sec x$

   ›Reveal solutionSolution

   This is a first-order linear ODE. Rewriting it in standard form and using an integrating factor gives $f(x) = (1-x)\sec x$, which matches option (A).

   We start with the given differential equation:

   $\cos x \frac{dy}{dx} = y \sin x - 1$

   The goal is to find $y = f(x)$ satisfying $f(0) = 1$, and then match it to one of the options.

   ---

   Concept and Intuition

   The equation is linear in $y$ but not yet in standard form. The standard form for a first-order linear ODE is:

   $\frac{dy}{dx} + P(x) y = Q(x)$

   Once in this form, we multiply through by an integrating factor $\mu(x) = e^{\int P(x)\,dx}$, which lets us write the left-hand side as the derivative of $\mu(x) y$. Then we integrate both sides.

   Here, dividing by $\cos x$ will give us $P(x) = -\tan x$ and $Q(x) = -\sec x$. The integrating factor simplifies nicely because $\int \tan x\,dx = -\log|\cos x|$, so $\mu(x) = \sec x$.

   ---

   Step-by-step solution

   1. Rewrite in standard form Divide both sides by $\cos x$ (valid since $x \neq (2n+1)\frac{\pi}{2}$):

   $\frac{dy}{dx} = y \tan x - \sec x$

   Bring the $y$ term to the left:

   $\frac{dy}{dx} - (\tan x) y = -\sec x$

   So $P(x) = -\tan x$ and $Q(x) = -\sec x$.

   2. Find the integrating factor

   $\mu(x) = e^{\int P(x)\,dx} = e^{\int -\tan x\,dx}$

   Since $\int \tan x\,dx = -\log|\cos x|$, we have:

   $\int -\tan x\,dx = \log|\cos x|$

   Hence:

   $\mu(x) = e^{\log|\cos x|} = |\cos x|$

   For the domain (where $\cos x > 0$ near $x=0$), we can take $\mu(x) = \cos x$. But it's more standard to use $\sec x$ as the integrating factor when we multiply through — let's check.

   Actually, careful: The standard formula is $\mu = e^{\int P\,dx}$. With $P = -\tan x$, we get $\mu = e^{\log(\cos x)} = \cos x$ (taking positive branch near 0). So the integrating factor is $\cos x$.

   3. Multiply the ODE by $\mu(x) = \cos x$ Original ODE in standard form:

   $\frac{dy}{dx} - (\tan x) y = -\sec x$

   Multiply by $\cos x$:

   $\cos x \frac{dy}{dx} - y \sin x = -1$

   Notice the left side is exactly $\frac{d}{dx}(y \cos x)$ because:

   $\frac{d}{dx}(y \cos x) = \frac{dy}{dx}\cos x - y\sin x$

   So we have:

   $\frac{d}{dx}(y \cos x) = -1$

   4. Integrate both sides

   $y \cos x = \int -1\,dx = -x + C$

   5. Apply the initial condition $f(0) = 1$ At $x=0$, $y=1$ and $\cos 0 = 1$:

   $1 \cdot 1 = -0 + C \quad\Rightarrow\quad C = 1$

   So:

   $y \cos x = 1 - x$

   Hence:

   $y = \frac{1-x}{\cos x} = (1-x)\sec x$

   6. Match with options This is exactly option (A).

   ---

   Tip

   A common mistake is to forget the sign when integrating $\tan x$ or to misplace the negative in the standard form. Always double-check that $\frac{d}{dx}(\mu y)$ matches after multiplying by $\mu$.

   Watch out

   The domain restriction $x \neq (2n+1)\frac{\pi}{2}$ ensures $\cos x \neq 0$, so division by $\cos x$ and the use of $\sec x$ are valid. At $x=0$, everything is well-defined.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

7. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.If $\dfrac{dy}{dx} - \dfrac{2x}{x^2 + b} y = -2x(x^2 + b),\ y(0) = 12,\ y(1) = 10$, then sum of all possible values of $b$ is (A) $1$ (B) $4$ (C) $-3$ (D) $-1$

   ›Reveal solutionSolution

   This is a first-order linear ODE solved via an integrating factor; the two boundary conditions force a specific value of the parameter $b$, and the sum of all possible $b$ values is $-3$.

   We are given the differential equation

   $\frac{dy}{dx} - \frac{2x}{x^2 + b}\, y = -2x(x^2 + b),$

   with conditions $y(0) = 12$ and $y(1) = 10$. The parameter $b$ is unknown, and we must find all possible $b$ that allow both conditions to hold, then sum them.

   ---

   Concept and intuition

   This is a first-order linear ODE of the form

   $\frac{dy}{dx} + P(x) y = Q(x).$

   The standard method: multiply by an integrating factor $\mu(x) = e^{\int P(x)\,dx}$ to make the left side a perfect derivative. Here $P(x) = -\frac{2x}{x^2+b}$, so the integrating factor will simplify nicely because the numerator is the derivative of the denominator. The right-hand side is a polynomial times $(x^2+b)$, so after multiplication we’ll integrate easily.

   The twist: we have two boundary conditions for a first-order ODE — that usually overdetermines the system. The parameter $b$ must adjust so that both conditions are consistent. We’ll solve the ODE in terms of $b$ and a constant $C$, then impose $y(0)=12$ and $y(1)=10$ to get equations that determine $b$.

   ---

   Step-by-step solution

   1. Identify $P(x)$ and compute the integrating factor.

   Rewrite the ODE as

   $\frac{dy}{dx} + \left(-\frac{2x}{x^2+b}\right) y = -2x(x^2+b).$

   So $P(x) = -\frac{2x}{x^2+b}$. Then

   $\int P(x)\,dx = -\int \frac{2x}{x^2+b}\,dx = -\log|x^2+b| + \text{constant}.$

   Thus the integrating factor is

   $\mu(x) = e^{\int P\,dx} = e^{-\log|x^2+b|} = \frac{1}{x^2+b}.$

   (We can drop absolute values since $b$ will be chosen so that $x^2+b>0$ on the interval containing 0 and 1, or we treat it as a formal algebraic factor.)

   2. Multiply the ODE by $\mu(x)$.

   $\frac{1}{x^2+b}\,\frac{dy}{dx} - \frac{2x}{(x^2+b)^2}\,y = -2x.$

   Notice the left side is exactly

   $\frac{d}{dx}\left( \frac{y}{x^2+b} \right).$

   Check: derivative of $\frac{y}{x^2+b}$ is $\frac{y'}{x^2+b} - \frac{2x y}{(x^2+b)^2}$. Yes.

   So we have

   $\frac{d}{dx}\left( \frac{y}{x^2+b} \right) = -2x.$

   3. Integrate both sides.

   $\frac{y}{x^2+b} = \int (-2x)\,dx = -x^2 + C,$

   where $C$ is an arbitrary constant.

   Thus

   $y(x) = (x^2+b)(-x^2 + C).$

   4. Apply the first condition $y(0)=12$.

   At $x=0$:

   $y(0) = (0+b)(0 + C) = bC = 12 \quad\Rightarrow\quad C = \frac{12}{b}.$

   5. Apply the second condition $y(1)=10$.

   At $x=1$:

   $y(1) = (1+b)(-1 + C) = 10.$

   Substitute $C = \frac{12}{b}$:

   $(1+b)\left(-1 + \frac{12}{b}\right) = 10.$

   6. Solve for $b$.

   Simplify the left side:

   $(1+b)\left(\frac{-b + 12}{b}\right) = \frac{(1+b)(12-b)}{b} = 10.$

   Multiply both sides by $b$ (note $b \neq 0$ because otherwise $y(0)=12$ would be impossible from $bC=12$):

   $(1+b)(12-b) = 10b.$

   Expand:

   $12 - b + 12b - b^2 = 10b \quad\Rightarrow\quad 12 + 11b - b^2 = 10b.$

   Bring all terms to one side:

   $12 + 11b - b^2 - 10b = 0 \quad\Rightarrow\quad 12 + b - b^2 = 0.$

   Multiply by $-1$:

   $b^2 - b - 12 = 0.$

   Factor:

   $(b-4)(b+3) = 0.$

   So $b = 4$ or $b = -3$.

   7. Sum all possible values of $b$.

   Sum = $4 + (-3) = 1$.

   Watch out

   A common mistake is to forget that $b$ cannot be zero (since $y(0)=bC$ would force $0=12$), but our quadratic already excludes $b=0$. Also, check that for $b=-3$, the denominator $x^2-3$ is nonzero at $x=0$ and $x=1$ (it is $-3$ and $-2$ respectively), so the ODE is well-defined on the interval.

   Tip

   The integrating factor method turned the ODE into a simple derivative because the coefficient $\frac{2x}{x^2+b}$ is exactly the derivative of $\log(x^2+b)$. This is a classic pattern: whenever $P(x)$ is a constant times $\frac{f'(x)}{f(x)}$, the integrating factor is a power of $f(x)$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

8. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.The general solution of the differential equation $(\sec x + \tan x) \dfrac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1$ is (A) $(1 + \sin x) y = n \cos x + c$ (B) $(1 + \cos x) y = x \sin x + c$ (C) $(\sec x + \tan x) y = x \sec x + c$ (D) $(\sec x + \tan x) y = x + c$

   ›Reveal solutionSolution

   This is a first-order linear ODE. Rewriting it in standard form and applying the integrating factor method shows that the general solution is $(\sec x + \tan x) y = x + c$, which matches option (D).

   The key concept is recognizing the equation as a first-order linear differential equation of the form

   $\frac{dy}{dx} + P(x) y = Q(x).$

   The standard method is to multiply through by an integrating factor $\mu(x) = e^{\int P\,dx}$, which makes the left side a perfect derivative. Here, the coefficients are cleverly arranged so that the integrating factor simplifies dramatically.

   1. Rewrite in standard form The given equation is

   $(\sec x + \tan x) \frac{dy}{dx} + (\sec^2 x + \sec x \tan x) y = 1.$

   Divide through by $(\sec x + \tan x)$ to isolate $\frac{dy}{dx}$:

   $\frac{dy}{dx} + \frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} \, y = \frac{1}{\sec x + \tan x}.$

   2. Simplify the coefficient of $y$ Factor the numerator: $\sec^2 x + \sec x \tan x = \sec x (\sec x + \tan x)$. Hence

   $\frac{\sec^2 x + \sec x \tan x}{\sec x + \tan x} = \sec x.$

   So the ODE becomes

   $\frac{dy}{dx} + (\sec x) y = \frac{1}{\sec x + \tan x}.$

   3. Find the integrating factor

   $\mu(x) = e^{\int \sec x \, dx}.$

   A standard integral: $\int \sec x \, dx = \log |\sec x + \tan x| + C$.

   Thus

   $\mu(x) = e^{\log |\sec x + \tan x|} = \sec x + \tan x.$

   (We take the positive branch for typical intervals.)

   4. Multiply through by $\mu(x)$

   $(\sec x + \tan x) \frac{dy}{dx} + (\sec x + \tan x)(\sec x) y = 1.$

   Notice the left side is exactly the derivative of $(\sec x + \tan x) y$ because

   $\frac{d}{dx} \big[ (\sec x + \tan x) y \big] = (\sec x + \tan x) \frac{dy}{dx} + (\sec x \tan x + \sec^2 x) y,$

   and indeed $\sec x (\sec x + \tan x) = \sec^2 x + \sec x \tan x$. So we have

   $\frac{d}{dx} \big[ (\sec x + \tan x) y \big] = 1.$

   5. Integrate both sides

   $(\sec x + \tan x) y = \int 1 \, dx = x + c.$

   6. Match with the options This is exactly option (D): $(\sec x + \tan x) y = x + c$.

   Tip

   Notice that the integrating factor turned out to be exactly the coefficient of $\frac{dy}{dx}$ in the original equation. This is a neat shortcut: if the ODE is already written as $\frac{d}{dx}[M(x) y] = \text{something}$, you can integrate directly without computing the integrating factor separately.

   Watch out

   A common mistake is to forget to divide by the coefficient of $\frac{dy}{dx}$ first, or to mis-simplify $\sec^2 x + \sec x \tan x$. Always check the algebra carefully.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $y = \sin x + A \cos x$ is the general solution of $\dfrac{dy}{dx} + f(x)y = \sec x$, then an integrating factor of the differential equation is (A) $\sec x$ (B) $\tan x$ (C) $\cos x$ (D) $\sin x$

   ›Reveal solutionSolution

   The given general solution is $y = \sin x + A \cos x$. Differentiating and substituting into the differential equation reveals $f(x) = \tan x$, so the integrating factor is $\sec x$, which is option (A).

   We are told that $y = \sin x + A \cos x$ (where $A$ is an arbitrary constant) is the general solution of

   $\frac{dy}{dx} + f(x) y = \sec x.$

   Our goal is to find the integrating factor (I.F.) of this linear first-order ODE. The standard form is $y' + P(x) y = Q(x)$, and the integrating factor is $\mu(x) = e^{\int P(x)\,dx}$. Here $P(x) = f(x)$.

   The key insight: If we already know the general solution, we can work backwards to find $f(x)$ by differentiating the solution and plugging it into the equation. Then we compute the integrating factor directly.

   ---

   1. Differentiate the given general solution

      $y = \sin x + A \cos x$

      $\frac{dy}{dx} = \cos x - A \sin x$

   2. Substitute into the differential equation

      The equation is $\frac{dy}{dx} + f(x) y = \sec x$.

      So:

   $(\cos x - A \sin x) + f(x)(\sin x + A \cos x) = \sec x$

   3. Group terms involving $A$ and those without $A$ Expand:

   $\cos x - A \sin x + f(x) \sin x + A f(x) \cos x = \sec x$

   Group constant (in $A$) terms: $\cos x + f(x) \sin x$

   Group $A$ terms: $A( -\sin x + f(x) \cos x )$

   Since this must hold for all $A$ (the solution is general), the coefficient of $A$ must be zero. That gives:

   $-\sin x + f(x) \cos x = 0 \quad \Rightarrow \quad f(x) \cos x = \sin x \quad \Rightarrow \quad f(x) = \tan x$

   4. Verify the constant part With $f(x) = \tan x$, the constant part becomes:

   $\cos x + \tan x \cdot \sin x = \cos x + \frac{\sin^2 x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x$

   which matches the right-hand side. So everything is consistent.

   5. Find the integrating factor For the linear ODE $y' + (\tan x) y = \sec x$, the integrating factor is

   $\mu(x) = e^{\int \tan x \, dx} = e^{\log|\sec x|} = \sec x$

   (ignoring the absolute value for typical domains).

   Tip

   A common pitfall is to assume the integrating factor is simply $e^{\int f(x) dx}$ without checking the sign or form. Here $f(x) = \tan x$ integrates to $\log|\sec x|$, giving $\sec x$ — not $\cos x$ or $\tan x$.

   Watch out

   If you mistakenly think the integrating factor is $e^{\int \sec x \, dx}$, you'd get something messy. Always identify $P(x)$ correctly from the standard form $y' + P(x)y = Q(x)$.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

10. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $l$ and $m$ are respectively the order and the degree of the differential equation $f(x)y'' + g(x)y' = \dfrac{4y}{x}$ whose general solution is $y = ax^2 + b \log x$, then $f(m) + g(m) =$ (A) $2l$ (B) $l$ (C) $3m$ (D) $1 + m$

    ›Reveal solutionSolution

    The key idea is to find the differential equation from its given general solution, then identify its order $l$ and degree $m$, and finally evaluate $f(m)+g(m)$ — the answer is $l$.

    We are told that the general solution of the differential equation

    $f(x)y'' + g(x)y' = \frac{4y}{x}$

    is $y = ax^2 + b \log x$, where $a$ and $b$ are arbitrary constants. The functions $f(x)$ and $g(x)$ are not given explicitly — they are to be determined from the fact that this $y$ satisfies the equation for all $a, b$.

    The problem asks for $f(m) + g(m)$, where $l$ is the order and $m$ is the degree of this differential equation. So we first need to find the differential equation itself.

    1. Find the derivatives of the given solution.

    $y = a x^2 + b \log x$

    Differentiate:

    $y' = 2a x + \frac{b}{x}$

    Differentiate again:

    $y'' = 2a - \frac{b}{x^2}$

    2. Eliminate the arbitrary constants $a$ and $b$.

       We have three equations: $y$, $y'$, $y''$ in terms of $a$ and $b$. We need one equation relating $y$, $y'$, $y''$ and $x$ alone — that is the differential equation.

       From $y'' = 2a - \frac{b}{x^2}$, we can solve for $a$ and $b$ in terms of $y''$ and something else. But a cleaner way:

       From $y' = 2a x + \frac{b}{x}$, multiply by $x$:

    $x y' = 2a x^2 + b$

    From $y = a x^2 + b \log x$, we have $a x^2 = y - b \log x$. Substitute into $x y'$:

    $x y' = 2(y - b \log x) + b = 2y - 2b \log x + b$

    This still has $b$. Instead, let's use $y''$ directly.

    From $y'' = 2a - \frac{b}{x^2}$, multiply by $x^2$:

    $x^2 y'' = 2a x^2 - b$

    But $2a x^2 = 2(y - b \log x)$ from $y = a x^2 + b \log x$. So:

    $x^2 y'' = 2y - 2b \log x - b$

    This still contains $b$. We need another relation to eliminate $b$.

    From $y' = 2a x + \frac{b}{x}$, multiply by $x$:

    $x y' = 2a x^2 + b$

    And $2a x^2 = 2(y - b \log x)$. So:

    $x y' = 2y - 2b \log x + b$

    Now subtract the $x^2 y''$ equation from this? Let's do it systematically.

    We have:

    $x y' = 2y - 2b \log x + b \quad \text{(1)}$

    $x^2 y'' = 2y - 2b \log x - b \quad \text{(2)}$

    Subtract (2) from (1):

    $x y' - x^2 y'' = (2y - 2b \log x + b) - (2y - 2b \log x - b) = 2b$

    So $b = \frac{1}{2}(x y' - x^2 y'')$.

    Now add (1) and (2):

    $x y' + x^2 y'' = (2y - 2b \log x + b) + (2y - 2b \log x - b) = 4y - 4b \log x$

    Substitute $b$:

    $x y' + x^2 y'' = 4y - 4 \left( \frac{1}{2}(x y' - x^2 y'') \right) \log x$

    $x y' + x^2 y'' = 4y - 2(x y' - x^2 y'') \log x$

    Bring terms together:

    $x y' + x^2 y'' + 2(x y' - x^2 y'') \log x = 4y$

    Factor:

    $x y' (1 + 2 \log x) + x^2 y'' (1 - 2 \log x) = 4y$

    Divide through by $x$ (assuming $x \neq 0$):

    $y' (1 + 2 \log x) + x y'' (1 - 2 \log x) = \frac{4y}{x}$

    This is the differential equation. Compare with the given form $f(x) y'' + g(x) y' = \frac{4y}{x}$:

    $f(x) = x(1 - 2 \log x), \quad g(x) = 1 + 2 \log x$

    3. Identify order $l$ and degree $m$.

       The highest derivative is $y''$, so order $l = 2$.

       The equation is polynomial in $y''$ and $y'$ (no fractional powers, no transcendental functions of derivatives), and the highest power of $y''$ is 1. So degree $m = 1$.

       Watch out

       Degree is defined only when the equation is polynomial in the derivatives. Here it is, so degree = 1. The presence of $\log x$ in the coefficients does not affect the degree — degree concerns only the dependent variable and its derivatives.

    4. Compute $f(m) + g(m)$.

       Since $m = 1$, evaluate $f(1)$ and $g(1)$:

    $f(1) = 1 \cdot (1 - 2 \log 1) = 1 \cdot (1 - 0) = 1$

    $g(1) = 1 + 2 \log 1 = 1 + 0 = 1$

    So $f(m) + g(m) = 1 + 1 = 2$.

    And $l = 2$, so $f(m) + g(m) = l$.

    ✓Final answer

    The value is $f(m) + g(m) = l$, which corresponds to option (B).

11. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
    Q.The general solution of $\dfrac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$ is (A) $\tan \dfrac{x}{2} = \dfrac{y^2}{2} - \cos y + C$ (B) $\tan \dfrac{y}{2} = \dfrac{x^2}{2} - \cos x + C$ (C) $\sec^2 \dfrac{y}{2} = \dfrac{x^2}{2} - \cos x + C$ (D) $\tan \dfrac{y}{2} = \dfrac{x^2}{2} + \cos x + Cx$

    ›Reveal solutionSolution

    The given differential equation is separable after factoring. The solution is found by integrating both sides, leading to $\tan\frac{y}{2} = \frac{x^2}{2} - \cos x + C$, which matches option (B).

    The key is to notice that the right-hand side can be grouped into terms that depend only on $x$ and terms that depend only on $y$. That’s the hallmark of a separable differential equation — and once you see the pattern, the integration is straightforward.

    Let’s rewrite the equation:

    $\frac{dy}{dx} = x + \sin x \cos y + x \cos y + \sin x$

    Group the terms cleverly:

    $\frac{dy}{dx} = (x + \sin x) + (\sin x \cos y + x \cos y)$

    Factor $\cos y$ from the last two terms:

    $\frac{dy}{dx} = (x + \sin x) + \cos y (x + \sin x)$

    Now factor $(x + \sin x)$ out of the whole right-hand side:

    $\frac{dy}{dx} = (x + \sin x)(1 + \cos y)$

    This is clearly separable: the $x$-part is $(x + \sin x)$ and the $y$-part is $(1 + \cos y)$.

    1. Separate the variables Bring all $y$ terms to the left and $x$ terms to the right:

    $\frac{dy}{1 + \cos y} = (x + \sin x)\, dx$

    2. Integrate both sides The left side uses a standard trigonometric identity. Recall:

    $1 + \cos y = 2\cos^2\frac{y}{2}$

    So:

    $\frac{1}{1 + \cos y} = \frac{1}{2\cos^2\frac{y}{2}} = \frac{1}{2}\sec^2\frac{y}{2}$

    Therefore:

    $\int \frac{dy}{1 + \cos y} = \frac{1}{2}\int \sec^2\frac{y}{2}\, dy$

    Let $u = y/2$, then $dy = 2\,du$, and:

    $\frac{1}{2}\int \sec^2 u \cdot 2\,du = \int \sec^2 u\, du = \tan u + C = \tan\frac{y}{2} + C$

    The right side integrates easily:

    $\int (x + \sin x)\, dx = \frac{x^2}{2} - \cos x + C$

    3. Combine the results Equating the two integrals (with a single constant of integration):

    $\tan\frac{y}{2} = \frac{x^2}{2} - \cos x + C$

    Watch out

    A common mistake is to forget the factor of $1/2$ when integrating $\sec^2(y/2)$. Always check the chain rule: the derivative of $\tan(y/2)$ is $\frac{1}{2}\sec^2(y/2)$, so the integral of $\sec^2(y/2)$ is $2\tan(y/2)$. Here the $1/2$ from the identity cancels that factor neatly.

    Tip

    The identity $1 + \cos y = 2\cos^2(y/2)$ is your best friend for integrals involving $1 + \cos y$ or $1 + \sin y$. Memorize it — it saves time in exams.

    ✓Final answer

    The correct option is (B): $\tan\frac{y}{2} = \frac{x^2}{2} - \cos x + C$.

12. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
    Q.If the system of linear equations $x + y + z = 1$, $2x + 2y + 3z = 6$, $x + 4y + 9z = 3$ has a unique solution $x = \alpha$, $y = \beta$, $z = \gamma$ then the value of $\beta$ is (A) $3$ (B) $6$ (C) $-10$ (D) $-4$

    ›Reveal solutionSolution

    The system is solved by elimination; the value of $\beta$ (y) is found to be $-10$, which corresponds to option (C).

    The key here is to recognize that this is a system of three linear equations in three unknowns. A unique solution exists when the equations are consistent and independent — meaning the planes intersect at a single point. We can solve it by systematic elimination, working step by step to isolate each variable.

    1. Write the system clearly:

    $\begin{aligned} x + y + z &= 1 \quad \text{(1)} \\ 2x + 2y + 3z &= 6 \quad \text{(2)} \\ x + 4y + 9z &= 3 \quad \text{(3)} \end{aligned}$

    2. Eliminate $x$ from equations (2) and (3) using equation (1). Subtract $2 \times$ (1) from (2):

    $(2x + 2y + 3z) - 2(x + y + z) = 6 - 2(1)$

    This simplifies to:

    $2x + 2y + 3z - 2x - 2y - 2z = 6 - 2$

    So:

    $z = 4$

    Tip

    Notice how the $x$ and $y$ terms cancel perfectly here — that’s because equation (2) is a multiple of (1) except for the $z$ term. This immediately gives $z$ without any further work.

    3. Now subtract equation (1) from equation (3) to eliminate $x$:

    $(x + 4y + 9z) - (x + y + z) = 3 - 1$

    This gives:

    $3y + 8z = 2$

    4. Substitute $z = 4$ into $3y + 8z = 2$:

    $3y + 8(4) = 2 \implies 3y + 32 = 2 \implies 3y = -30 \implies y = -10$

    So $\beta = y = -10$.

    5. (Optional check) We could find $x$ from equation (1): $x + (-10) + 4 = 1 \implies x - 6 = 1 \implies x = 7$. The solution $(7, -10, 4)$ indeed satisfies all three equations, confirming uniqueness.

    Watch out

    A common mistake is to try Cramer’s rule or matrix inversion without first checking if elimination is simpler. Here, the coefficients are designed so that elimination gives $z$ instantly — don’t overcomplicate.

    ✓Final answer

    The value of $\beta$ is $\boxed{-10}$, which is option (C).