Q.Verify that the function $y = a\cos x + b\sin x$, where $a, b \in \mathbf{R}$ is a solution of the differential equation $\frac{d^2y}{dx^2} + y = 0$.

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

   🔒 Start your 14-day free trial to see the full solution →
2. 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:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
3. 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

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

   🔒 Start your 14-day free trial to see the full solution →
4. 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:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
5. 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: $\cos x + \tan x \cdot \sin x = \cos x + \frac{\sin^2 x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\cos x} = \frac{1}{\cos x} = \sec x$ …

   🔒 Start your 14-day free trial to see the full solution →
6. 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:

   $$$$

   🔒 Start your 14-day free trial to see the full solution →
7. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.If $y = \left( x + \sqrt{x^2 + 1} \right)^5$ then $25y =$ (A) $(x^2 + 1)y_2 - x y_1$ (B) $(x^2 + 1)y_2 + x y_1$ (C) $(x^2 + 1)y_2 - 2x y_1$ (D) $(x^2 + 1)y_2 + 2x y_1$

   ›Reveal solutionSolution

   The function $y = (x + \sqrt{x^2+1})^5$ is a power of an inverse hyperbolic sine, so its derivatives satisfy a simple recurrence. Differentiating twice and rearranging gives $25y = (x^2+1)y_2 + x y_1$, which is option (B).

   We have $y = \left( x + \sqrt{x^2 + 1} \right)^5$. The expression inside the parentheses is the standard form for the inverse hyperbolic sine: $\sinh^{-1} x = \log(x + \sqrt{x^2+1})$, so $x + \sqrt{x^2+1} = e^{\sinh^{-1} x}$. That means $y = e^{5\sinh^{-1} x}$. This is a composition that makes differentiation clean: the derivative of $\sinh^{-1} x$ is $\frac{1}{\sqrt{x^2+1}}$, and the chain rule will produce a pattern that eliminates the square root.

   The key insight: instead of brute-force expanding, we can find a relation between $y$, $y_1 = \frac{dy}{dx}$, and $y_2 = \frac{d^2y}{dx^2}$ by differentiating the defining equation. Notice that $x + \sqrt{x^2+1}$ satisfies a neat property: its reciprocal is $\sqrt{x^2+1} - x$. This will help us isolate derivatives.

   1. First derivative. Let $u = x + \sqrt{x^2+1}$. Then $y = u^5$, and $\frac{du}{dx} = 1 + \frac{x}{\sqrt{x^2+1}} = \frac{\sqrt{x^2+1} + x}{\sqrt{x^2+1}} = \frac{u}{\sqrt{x^2+1}}$. So by the chain rule:

   $y_1 = 5u^4 \cdot \frac{du}{dx} = 5u^4 \cdot \frac{u}{\sqrt{x^2+1}} = \frac{5u^5}{\sqrt{x^2+1}} = \frac{5y}{\sqrt{x^2+1}}.$

   Hence

   $\sqrt{x^2+1}\, y_1 = 5y. \qquad(1)$

   2. Second derivative. Differentiate (1) with respect to $x$. The left side is a product:

   $\frac{d}{dx}\left( \sqrt{x^2+1}\, y_1 \right) = \frac{x}{\sqrt{x^2+1}}\, y_1 + \sqrt{x^2+1}\, y_2.$

   The right side differentiates to $5y_1$. So: …

   🔒 Start your 14-day free trial to see the full solution →
8. 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:

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

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $-3$:

   $\log|u| = -3\log|x| - 3C$ …

   🔒 Start your 14-day free trial to see the full solution →
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:

    $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
    Q.If the equation of the line joining the points $A(x_1, y_1)$ and $B(x_2, y_2)$ is $ax + by = c$ and the distance between A and B is $\sqrt{a^2 + b^2}$, then $c^2 =$ (A) $\dfrac{x_1^2 + y_1^2}{x_2 y_2}$ (B) $\dfrac{x_1 + x_2}{y_1 + y_2}$ (C) $(x_1 y_2 - x_2 y_1)^2$ (D) $(x_1 x_2 - y_1 y_2)^2$

    ›Reveal solutionSolution

    The key idea is that the distance condition forces the line’s normal vector $(a,b)$ to be exactly the vector from $A$ to $B$, so the line equation becomes $(x_2-x_1)x + (y_2-y_1)y = c$, and plugging in either point gives $c = x_1x_2 - x_1^2 + y_1y_2 - y_1^2$, which simplifies to $(x_1y_2 - x_2y_1)^2$ after using the distance condition. The correct option is (C).

    ---

    Concept and intuition

    We are told that the line through $A(x_1,y_1)$ and $B(x_2,y_2)$ has equation $ax+by=c$, and that the distance $AB = \sqrt{a^2+b^2}$.

    Normally, the distance between two points is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The fact that this equals $\sqrt{a^2+b^2}$ suggests that $(a,b)$ is exactly the vector from $A$ to $B$ (or its negative). Because if the line’s normal vector $(a,b)$ has the same length as the segment $AB$, and the line passes through both points, then the normal must be aligned with the direction of the segment — a very special condition. This lets us identify $a$ and $b$ up to sign, then find $c$ by substituting either point.

    ---

    Step-by-step reasoning

    1. Write the distance condition The distance between $A$ and $B$ is

    $AB = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.$

    We are given $AB = \sqrt{a^2+b^2}$, so

    $(x_2-x_1)^2 + (y_2-y_1)^2 = a^2 + b^2. \qquad(1)$

    2. Use that both points lie on the line Since $A$ and $B$ satisfy $ax+by=c$, we have

    $a x_1 + b y_1 = c \quad\text{and}\quad a x_2 + b y_2 = c. \qquad(2)$

    Subtracting these gives

    $a(x_2-x_1) + b(y_2-y_1) = 0. \qquad(3)$

    3. Interpret equation (3)

       Equation (3) says the vector $(a,b)$ is perpendicular to the vector $(x_2-x_1,\; y_2-y_1)$. But from (1) their lengths are equal. Two perpendicular vectors of equal length? That forces them to be rotated by 90° relative to each other. In the plane, if a vector $(u,v)$ has length $L$, then a perpendicular vector of the same length is either $(-v, u)$ or $(v, -u)$.

       So we must have

    $(a,b) = \pm\,( -(y_2-y_1),\; x_2-x_1 ).$

    (Check: dot product $(x_2-x_1)(-(y_2-y_1)) + (y_2-y_1)(x_2-x_1) = 0$, and lengths match.)

    4. Find $c$ using one point Take the positive sign (the negative sign will give the same $c^2$). Then

    $a = -(y_2-y_1), \quad b = x_2-x_1.$

    Plug into $a x_1 + b y_1 = c$:

    $$$$

    🔒 Start your 14-day free trial to see the full solution →
12. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
    Q.If the point $(x, y)$ satisfies the equation $\dfrac{x + i(x-2)}{3 + i} - i = \dfrac{2y + i(1-3y)}{i-3}$, then $x + y =$ (A) 4 (B) 2 (C) 0 (D) -2

    ›Reveal solutionSolution

    The key idea is to simplify the given complex equation by rationalising denominators and equating real and imaginary parts, leading to $x + y = 2$.

    The problem gives an equation involving complex numbers, where $x$ and $y$ are real numbers. Our goal is to find $x + y$. The approach is to treat the equation as an equality of two complex numbers, simplify each side separately, and then equate the real and imaginary parts.

    Let’s work through it step by step.

    1. Write the given equation clearly The equation is:

    $\frac{x + i(x-2)}{3 + i} - i = \frac{2y + i(1-3y)}{i-3}$

    Notice that $i-3 = -(3-i)$, which will be useful later.

    2. Simplify the left-hand side (LHS) First, combine the terms on LHS:

    $\text{LHS} = \frac{x + i(x-2)}{3 + i} - i$

    Write $-i$ as a fraction with denominator $1$:

    $\text{LHS} = \frac{x + i(x-2)}{3 + i} - \frac{i(3+i)}{3+i} = \frac{x + i(x-2) - i(3+i)}{3+i}$

    Simplify the numerator:

    $x + i(x-2) - i(3+i) = x + i(x-2) - 3i - i^2$

    Since $i^2 = -1$, we have $-i^2 = -(-1) = 1$? Wait carefully: $- i(3+i) = -3i - i^2 = -3i - (-1) = -3i + 1$. So:

    $\text{Numerator} = x + i(x-2) - 3i + 1 = (x+1) + i[(x-2) - 3] = (x+1) + i(x-5)$

    Thus:

    $\text{LHS} = \frac{(x+1) + i(x-5)}{3+i}$

    3. Simplify the right-hand side (RHS) The RHS is:

    $\frac{2y + i(1-3y)}{i-3}$

    Since $i-3 = -(3-i)$, we can write:

    $\text{RHS} = \frac{2y + i(1-3y)}{-(3-i)} = -\frac{2y + i(1-3y)}{3-i}$

    Alternatively, we could rationalise directly. Let’s keep it as is for now.

    4. Rationalise both sides Multiply numerator and denominator of LHS by the conjugate of $3+i$, which is $3-i$:

    $\text{LHS} = \frac{[(x+1) + i(x-5)](3-i)}{(3+i)(3-i)} = \frac{[(x+1) + i(x-5)](3-i)}{9 - i^2} = \frac{[(x+1) + i(x-5)](3-i)}{10}$

    Expand the numerator:

    $(x+1)(3-i) + i(x-5)(3-i) = 3(x+1) - i(x+1) + 3i(x-5) - i^2(x-5)$

    Since $i^2 = -1$, $-i^2(x-5) = (x-5)$. So:

    $= 3x+3 - i(x+1) + 3i x - 15i + x - 5$

    Combine real parts: $3x + 3 + x - 5 = 4x - 2$

    Combine imaginary parts: $-i(x+1) + 3i x - 15i = i(-x -1 + 3x - 15) = i(2x - 16)$

    Thus:

    $\text{LHS} = \frac{(4x - 2) + i(2x - 16)}{10}$

    Now rationalise RHS. We have:

    $\text{RHS} = -\frac{2y + i(1-3y)}{3-i}$

    Multiply numerator and denominator of the fraction by the conjugate $3+i$:

    $\frac{2y + i(1-3y)}{3-i} = \frac{[2y + i(1-3y)](3+i)}{(3-i)(3+i)} = \frac{[2y + i(1-3y)](3+i)}{10}$

    Expand the numerator:

    $2y(3+i) + i(1-3y)(3+i) = 6y + 2iy + 3i(1-3y) + i^2(1-3y)$

    Since $i^2 = -1$, the last term becomes $-(1-3y) = -1 + 3y$. So:

    $= 6y + 2iy + 3i - 9iy - 1 + 3y$

    Combine real parts: $6y - 1 + 3y = 9y - 1$

    Combine imaginary parts: $2iy + 3i - 9iy = i(2y + 3 - 9y) = i(3 - 7y)$

    Thus:

    $\frac{2y + i(1-3y)}{3-i} = \frac{(9y - 1) + i(3 - 7y)}{10}$

    Therefore:

    $$$$

    🔒 Start your 14-day free trial to see the full solution →