Q.The value of ''n , such that the differential equation 𝒙𝒏 𝒅𝒚 𝒅𝒙 = 𝒚(𝒍𝒐𝒈𝒚 − 𝒍𝒐𝒈𝒙 + 𝟏); (𝐰𝐡𝐞𝐫𝐞 𝒙, 𝒚 ∈ 𝑹+) is homogeneous, is
(A) 0
(B) 1
(C) 2
(D) 3

Showing the 12 most recent of 13 on this concept.

1. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.Assertion (A) : The degree of the differential equation $y'' + 2xy' + \log_e\left(\dfrac{dy}{dx}\right) = 0$ is 2. Reason (R) : The degree of a differential equation is the highest degree of the highest order derivative occurring in the equation, after the equation is expressed in the form of a polynomial in differential coefficients. The correct option among the following is: (A) (A) is true (R) is true and (R) is the correct explanation for (A) (B) (A) is true (R) is true but (R) is not the correct explanation for (A) (C) (A) is true but (R) is false (D) (A) is false but (R) is true

   ›Reveal solutionSolution

   Because of the $\log_e(dy/dx)$ term the equation is not a polynomial in its derivatives, so its degree is not defined — the Assertion (degree $=2$) is false. The Reason states the definition correctly and is true. Option (D).

   The concept: order versus degree

   • Order = the order of the highest derivative present. Here the highest is $y''$, so the order is 2.
   • Degree = the power to which that highest-order derivative is raised — but only after the equation has been written as a polynomial in the derivatives. If it cannot be so written, the degree is not defined.

   The degree is undefined whenever a derivative appears inside a transcendental function: $\sin(y')$, $e^{y'}$, $\log(y')$, and so on.

   Step 1 — Look at the equation

   $y'' + 2x\,y' + \log_e\!\left(\frac{dy}{dx}\right) = 0$

   The first two terms are fine, but $\log_e(y')$ is a transcendental function of $y'$. Expanding it as an infinite series would give infinitely many powers of $y'$ — never a polynomial. There is no algebraic manipulation that removes the logarithm while keeping the equation polynomial in $y'$ and $y''$.

   Step 2 — Verdict on the Assertion …

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2. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If the solution for the differential equation $y^2 dx + (x^2 - xy - y^2) dy = 0$ at $(2,1)$ is $x + y = k(xy^2 - y^3)$, then $k =$ (A) $-3$ (B) $-4$ (C) $4$ (D) $3$

   ›Reveal solutionSolution

   This is a homogeneous differential equation solved by the substitution $y = vx$. After separating variables and integrating, the general solution is $x + y = C(xy^2 - y^3)$. Using the point $(2,1)$ gives $C = -3$, so $k = -3$.

   We are given the differential equation

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

   and told that its solution passing through $(2,1)$ can be written as

   $x + y = k(xy^2 - y^3).$

   We need to find $k$.

   ---

   Concept and intuition

   The equation is homogeneous — every term in $dx$ and $dy$ is of degree 2. For such equations, the standard trick is to set $y = vx$ (or $x = vy$), which reduces the problem to a separable one. Once we integrate, we get a family of curves. Plugging the given point pins down the constant, and matching the form reveals $k$.

   ---

   Step-by-step solution

   1. Rewrite the equation in standard form

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

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

   $y^2 \frac{dx}{dy} + x^2 - xy - y^2 = 0$

   So

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

   The right-hand side is homogeneous of degree 0 (each term in numerator and denominator is degree 2). This suggests the substitution $x = vy$.

   2. Substitute $x = vy$ Then $\frac{dx}{dy} = v + y\frac{dv}{dy}$. Plug into the equation:

   $v + y\frac{dv}{dy} = \frac{-(vy)^2 + (vy)y + y^2}{y^2} = \frac{-v^2 y^2 + v y^2 + y^2}{y^2} = -v^2 + v + 1.$

   So

   $y\frac{dv}{dy} = -v^2 + v + 1 - v = -v^2 + 1.$

   3. Separate variables

   $y\frac{dv}{dy} = 1 - v^2 \quad\Rightarrow\quad \frac{dv}{1 - v^2} = \frac{dy}{y}.$

   4. Integrate both sides

   $\int \frac{dv}{1 - v^2} = \int \frac{dy}{y}.$

   The left-hand side is a standard partial fractions integral:

   $\frac{1}{1 - v^2} = \frac{1}{2}\left(\frac{1}{1-v} + \frac{1}{1+v}\right),$

   so

   $\int \frac{dv}{1 - v^2} = \frac{1}{2}\log\left|\frac{1+v}{1-v}\right|.$

   The right-hand side gives $\log|y| + C$. Thus

   $\frac{1}{2}\log\left|\frac{1+v}{1-v}\right| = \log|y| + C.$

   5. Simplify the constant Multiply by 2:

   $\log\left|\frac{1+v}{1-v}\right| = 2\log|y| + 2C = \log(y^2) + \log C_1,$

   where $C_1 = e^{2C} > 0$. So

   $\log\left|\frac{1+v}{1-v}\right| = \log(C_1 y^2) \quad\Rightarrow\quad \frac{1+v}{1-v} = C_1 y^2.$

   6. Back-substitute $v = x/y$ $$\frac{1 + x/y}{1 - x/y} = C_1 y^2 \quad\Rightarrow\quad …$$

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.The solution of the differential equation $\dfrac{dy}{dx} = \dfrac{3^{x+y} - 2 \cdot 3^x}{3^{x+y} - 2 \cdot 3^y}$ when $y(1) = 2$ is (A) $3^y = 7(3^x) + 12$ (B) $y = \log_3 \left( 7(3^x) - 14 \right)$ (C) $y = \log_3 \left( 7(3^x) - 12 \right)$ (D) $3^y = 7(3^x) - 14$

   ›Reveal solutionSolution

   This problem involves solving a first-order differential equation by separating variables. We simplify the expression, integrate both sides, and then use the given initial condition to find the particular solution. The solution is $y = \log_3 \left( 7(3^x) - 12 \right)$.

   A differential equation relates a function to its derivatives. To solve such an equation, we aim to find the original function. One common and powerful technique for first-order differential equations is separation of variables. This method works when the equation can be rearranged such that all terms involving the dependent variable (here, $y$) and its differential ($dy$) are on one side, and all terms involving the independent variable (here, $x$) and its differential ($dx$) are on the other side. Once separated, we can integrate both sides independently to find the general solution. Finally, any given initial condition allows us to determine the specific constant of integration, leading to a particular solution.

   Here's how we solve the given differential equation:

   1. Simplify the expression: The given differential equation is $\dfrac{dy}{dx} = \dfrac{3^{x+y} - 2 \cdot 3^x}{3^{x+y} - 2 \cdot 3^y}$. We can use the property $a^{m+n} = a^m \cdot a^n$ to rewrite $3^{x+y}$ as $3^x \cdot 3^y$.

   $\dfrac{dy}{dx} = \dfrac{3^x \cdot 3^y - 2 \cdot 3^x}{3^x \cdot 3^y - 2 \cdot 3^y}$

   Now, factor out common terms from the numerator and the denominator:
   

   $\dfrac{dy}{dx} = \dfrac{3^x(3^y - 2)}{3^y(3^x - 2)}$

   2. Separate the variables: Our goal is to get all terms involving $y$ on the left side with $dy$, and all terms involving $x$ on the right side with $dx$. Multiply both sides by $3^y(3^x - 2)$ and by $dx$:

   $\dfrac{3^y}{3^y - 2} dy = \dfrac{3^x}{3^x - 2} dx$

   The variables are now successfully separated.
   

   3. Integrate both sides:

   We integrate both sides of the separated equation:

   $\int \dfrac{3^y}{3^y - 2} dy = \int \dfrac{3^x}{3^x - 2} dx$

   To solve these integrals, we use a substitution. Consider the integral $\int \dfrac{3^t}{3^t - 2} dt$.
   Let $u = 3^t - 2$.
   Then, differentiate $u$ with respect to $t$: $du = \dfrac{d}{dt}(3^t - 2) dt = (3^t \ln 3) dt$.
   This means $3^t dt = \dfrac{du}{\ln 3}$.
   Substituting these into the integral:
   

   $\int \dfrac{1}{u} \cdot \dfrac{du}{\ln 3} = \dfrac{1}{\ln 3} \int \dfrac{1}{u} du = \dfrac{1}{\ln 3} \ln|u| + C$

   Replacing $u$ with $3^t - 2$:
   

   $\int \dfrac{3^t}{3^t - 2} dt = \dfrac{1}{\ln 3} \ln|3^t - 2| + C$

   Applying this result to both sides of our separated differential equation:
   

   $\dfrac{1}{\ln 3} \ln|3^y - 2| = \dfrac{1}{\ln 3} \ln|3^x - 2| + C'$

   Here, $C'$ is the constant of integration. We can multiply the entire equation by $\ln 3$ to simplify:
   

   $\ln|3^y - 2| = \ln|3^x - 2| + C''$

   Let $C'' = \ln K$ for some positive constant $K$.
   

   $\ln|3^y - 2| = \ln|3^x - 2| + \ln K$

   Using the logarithm property $\ln a + \ln b = \ln(ab)$: …
   

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If the general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{2x + 2y - 1}{x + y - 5} = 0$ is $ax + by - 9\log(|x + y + p|) = c$, and $b, a, p$ are in GP, then the common ratio of this GP is (A) $3$ (B) $2$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{3}$

   ›Reveal solutionSolution

   The differential equation is reducible to a homogeneous form via substitution $u = x + y$, leading to a solution of the form $ax + by - 9\log(|x + y + p|) = c$. Matching coefficients and using the GP condition $b, a, p$ gives the common ratio $2$.

   ---

   The key insight is that the equation mixes $x$ and $y$ in the combination $x + y$ in both numerator and denominator. This suggests a substitution that simplifies the structure: let $u = x + y$. Then the equation becomes separable, and after integration we obtain a logarithmic relation. Comparing with the given general form lets us identify $a$, $b$, and $p$, and the geometric progression condition then fixes the common ratio.

   ---

   1. Rewrite the equation in terms of $u = x + y$. Let $u = x + y$. Then $\frac{dy}{dx} = \frac{du}{dx} - 1$. Substitute into

   $\frac{dy}{dx} + \frac{2x + 2y - 1}{x + y - 5} = 0$

   to get

   $\left(\frac{du}{dx} - 1\right) + \frac{2u - 1}{u - 5} = 0.$

   2. Simplify to a separable form. Bring terms together:

   $\frac{du}{dx} = 1 - \frac{2u - 1}{u - 5}.$

   Compute the right-hand side:

   $1 - \frac{2u - 1}{u - 5} = \frac{(u - 5) - (2u - 1)}{u - 5} = \frac{-u - 4}{u - 5}.$

   So

   $\frac{du}{dx} = -\frac{u + 4}{u - 5}.$

   3. Separate variables and integrate.

   $\frac{u - 5}{u + 4} \, du = -dx.$

   Perform polynomial division:

   $\frac{u - 5}{u + 4} = 1 - \frac{9}{u + 4}.$

   Hence

   $\left(1 - \frac{9}{u + 4}\right) du = -dx.$

   Integrate:

   $\int \left(1 - \frac{9}{u + 4}\right) du = -\int dx$

   gives

   $u - 9 \log|u + 4| = -x + C.$

   4. Return to $x, y$ and match the given form. Since $u = x + y$, we have

   $x + y - 9 \log|x + y + 4| = -x + C.$

   Bring $x$ to the left:

   $2x + y - 9 \log|x + y + 4| = C.$ …

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5. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.The degree of the differential equation $x \left( \dfrac{d^2 y}{dx^2} \right)^{1/3} + 2x^2 \left( \dfrac{d^2 y}{dx^2} \right)^{5/3} + 7 \dfrac{dy}{dx} + y = 0$ (A) 15 (B) 5 (C) 12 (D) 3

   ›Reveal solutionSolution

   The degree of a differential equation is the power of the highest-order derivative once the equation is rewritten as a polynomial in the derivatives (no fractional or negative powers on any derivative). Here the highest-order derivative is $\frac{d^2y}{dx^2}$, and after the fractional exponents $1/3$ and $5/3$ are cleared, its highest surviving power is 5. So the degree is 5, option (B).

   ---

   Concept and intuition

   The degree of a differential equation is defined only once the equation is free of radicals and fractional/negative powers of the derivatives — it is then the exponent of the highest-order derivative present. When a derivative appears with more than one fractional power (as $\frac{d^2y}{dx^2}$ does here, with exponents $1/3$ and $5/3$), we clear the fractions by substitution and algebraic elimination, then read off the resulting integer power.

   ---

   Step-by-step solution

   1. Identify the highest-order derivative and its exponents

   $x \left( \frac{d^2y}{dx^2} \right)^{1/3} + 2x^2 \left( \frac{d^2y}{dx^2} \right)^{5/3} + 7\frac{dy}{dx} + y = 0.$

   The highest-order derivative is $p = \dfrac{d^2y}{dx^2}$ (order 2), appearing with exponents $\tfrac13$ and $\tfrac53$ — both with denominator 3.

   2. Substitute to remove the fractional exponents Let $t = p^{1/3}$ (so $p = t^3$) and $Q = 7\dfrac{dy}{dx} + y$. The equation becomes

   $2x^2 t^5 + x t + Q = 0,$

   which is already a genuine polynomial — but in $t$, not yet in $p$.

   3. Eliminate $t$ to get a polynomial purely in $p$ Since $t$ is a real cube root of $p$, the two other cube roots $t\omega,\ t\omega^2$ (where $\omega$ is a complex cube root of unity, $\omega^3=1$) also satisfy $(t\omega)^3=(t\omega^2)^3=p$. Multiplying the equation evaluated at $t,\ t\omega,\ t\omega^2$ together produces an expression that depends only on $p$ (it is symmetric in the three cube roots), via the identity $a^3+b^3+c^3-3abc = (a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)$ with $a = 2x^2p^{5/3},\ b = xp^{1/3},\ c=Q$: $8x^6p^5 - 6x^3Qp^2 + x^3p + Q^3 = 0.$ …

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6. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.If a and b are the arbitrary constants, then the differential equation corresponding to the family of curves given by $y = x \left[ a \cos(\log x) + b \sin(\log x) \right]$ is (A) $x \dfrac{d^2 y}{dx^2} + x \dfrac{dy}{dx} - 2y = 0$ (B) $x^2 \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} + 2y = 0$ (C) $x^2 \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} - 2y = 0$ (D) $x^2 \dfrac{d^2 y}{dx^2} - x \dfrac{dy}{dx} + y = 0$

   ›Reveal solutionSolution

   The family of curves is $x$ times a linear combination of $\cos(\log x)$ and $\sin(\log x)$, so it is the general solution of an Euler-Cauchy equation. Differentiating twice and eliminating $a$ and $b$ yields $x^2 y'' - x y' + 2y = 0$, which is option (B).

   We are given the family of curves

   $y = x \bigl[ a \cos(\log x) + b \sin(\log x) \bigr],$

   where $a$ and $b$ are arbitrary constants. The task is to find the differential equation (DE) that all such curves satisfy, without the constants $a$ and $b$.

   Concept and intuition:

   Since $a$ and $b$ are the only free parameters, the DE must be of second order (two constants → two derivatives to eliminate them). The presence of $\log x$ inside trigonometric functions, multiplied by $x$, is the classic signature of an Euler–Cauchy equation — terms like $x^2 y''$, $x y'$, and $y$ are the natural building blocks. Our plan: differentiate twice, then algebraically eliminate $a$ and $b$.

   ---

   1. First derivative Write $y = x \cdot u$, where $u = a \cos(\log x) + b \sin(\log x)$. Differentiate using the product rule:

   $y' = u + x u'.$

   Now $u' = \frac{d}{dx} \bigl[ a \cos(\log x) + b \sin(\log x) \bigr]$. Recall $\frac{d}{dx} \cos(\log x) = -\sin(\log x) \cdot \frac{1}{x}$, and $\frac{d}{dx} \sin(\log x) = \cos(\log x) \cdot \frac{1}{x}$. Hence

   $u' = \frac{1}{x} \bigl[ -a \sin(\log x) + b \cos(\log x) \bigr] = \frac{v}{x}, \quad \text{where } v = -a \sin(\log x) + b \cos(\log x).$

   So

   $y' = u + x \cdot \frac{v}{x} = u + v.$

   2. Second derivative Differentiate $y' = u + v$ again. We already have $u' = v/x$. Differentiating $v$:

   $v' = \frac{d}{dx}\bigl[-a\sin(\log x) + b\cos(\log x)\bigr] = \frac{1}{x}\bigl[-a\cos(\log x) - b\sin(\log x)\bigr] = -\frac{u}{x}.$

   So

   $y'' = u' + v' = \frac{v}{x} - \frac{u}{x} = \frac{v - u}{x}.$

   3. Eliminate $a$ and $b$ From $y = xu$, we get $u = \dfrac{y}{x}$. From $y' = u + v$, we get $v = y' - u = y' - \dfrac{y}{x}$. Substitute both into $y'' = \dfrac{v-u}{x}$: $$$$

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7. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.The general solution of the differential equation $(3x-4y)(dx-3dy)+(6dx-4dy)=0$ is (A) $x-2y+\log|3x-4y+6|=c$ (B) $5x-15y-4\log|15x-20y-12|=c$ (C) $5x-15y+14\log|15x-20y-12|=c$ (D) $8y-4x+\log|9x-12y+4|=c$

   ›Reveal solutionSolution

   Collect $dx$ and $dy$, substitute $v=3x-4y$; the solution is $5x-15y+14\log|15x-20y-12|=c$.

   Collect terms. Expanding $(3x-4y)(dx-3dy)+(6dx-4dy)=0$:

   $(3x-4y+6)\,dx + \big(-9x+12y-4\big)\,dy = 0 \quad\Rightarrow\quad \frac{dy}{dx}=\frac{3x-4y+6}{9x-12y+4}.$

   Substitution. Since $9x-12y = 3(3x-4y)$, let $v = 3x-4y$, giving $\dfrac{dy}{dx}=\dfrac{3-\frac{dv}{dx}}{4}$ and $\dfrac{dy}{dx}=\dfrac{v+6}{3v+4}$. Equating:

   $\frac{3-\frac{dv}{dx}}{4}=\frac{v+6}{3v+4}\ \Rightarrow\ \frac{dv}{dx}=3-\frac{4(v+6)}{3v+4}=\frac{5v-12}{3v+4}.$

   Separate and integrate. $\displaystyle\int\frac{3v+4}{5v-12}\,dv=\int dx$. Writing $3v+4=\tfrac{3}{5}(5v-12)+\tfrac{56}{5}$: …

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8. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.The general solution of the differential equation $(6x^2-2xy-18x+3y)dx-(x^2-3x)dy=0$ is (A) $2x^2-x^2y-9x^2+3xy+c=0$ (B) $4x^3-2x^2y-6x^2+6xy+c=0$ (C) $2x^2-4xy-y^2-x+3y+c=0$ (D) $3x^2+5xy-2y^2-4x-2y+c=0$

   ›Reveal solutionSolution

   The differential equation is exact after rewriting, and integrating yields $2x^3 - x^2y - 9x^2 + 3xy = c$, which matches option (A).

   We are given:

   $(6x^2-2xy-18x+3y)\,dx - (x^2-3x)\,dy = 0$

   ---

   Concept & Intuition

   This is a first-order differential equation. The standard approach is to check if it is exact. An equation of the form $M\,dx + N\,dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. If exact, we can find a function $F(x,y)$ whose total differential equals $M\,dx + N\,dy$; then the solution is $F(x,y) = c$.

   Here, note the minus sign: we have $M\,dx + N\,dy = 0$ with $M = 6x^2 - 2xy - 18x + 3y$ and $N = -(x^2 - 3x) = -x^2 + 3x$.

   ---

   Step-by-step solution

   1. Identify $M$ and $N$ Write the equation as:

   $(6x^2 - 2xy - 18x + 3y)\,dx + (-x^2 + 3x)\,dy = 0$

   So $M = 6x^2 - 2xy - 18x + 3y$ and $N = -x^2 + 3x$.

   2. Check exactness Compute $\frac{\partial M}{\partial y}$:

   $\frac{\partial M}{\partial y} = -2x + 3$

   Compute $\frac{\partial N}{\partial x}$:

   $\frac{\partial N}{\partial x} = -2x + 3$

   They are equal, so the equation is exact.

   3. Find the potential function $F(x,y)$ We need $F$ such that:

   $\frac{\partial F}{\partial x} = M = 6x^2 - 2xy - 18x + 3y$

   Integrate with respect to $x$:

   $F(x,y) = \int (6x^2 - 2xy - 18x + 3y)\,dx = 2x^3 - x^2y - 9x^2 + 3xy + g(y)$

   where $g(y)$ is an arbitrary function of $y$ alone.

   4. Determine $g(y)$ Differentiate $F$ with respect to $y$: $$\frac{\partial F}{\partial y} = -x^2 + 3x + g'(y) …$$

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9. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.The equation of any member of the family of all the ellipses whose axes are along the coordinate axes satisfies the differential equation (A) $xyy'' + x(y')^2 - y' = 0$ (B) $xyy'' + x(y')^2 - y = y'$ (C) $y'' + \dfrac{(y')^2}{y} - \dfrac{y}{x} = 0$ (D) $y'' + (y')^2 + x^2 y^2 = 0$

   ›Reveal solutionSolution

   The family of ellipses with axes along the coordinate axes has the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Eliminating the two arbitrary constants $a$ and $b$ by differentiating twice yields the differential equation $xyy'' + x(y')^2 - yy' = 0$, which matches option (A) after a sign check.

   The key idea: a family of curves with two independent parameters (here $a$ and $b$) requires two derivatives to eliminate them. The resulting differential equation must be free of both constants. We start from the standard ellipse equation and differentiate implicitly, then eliminate $a^2$ and $b^2$ algebraically.

   1. Write the general ellipse equation. Any ellipse with axes along the coordinate axes (centre at the origin) has the form

   $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,$

   where $a$ and $b$ are positive constants (semi-major and semi-minor axes). No other parameters appear, so the family is two-parameter.

   2. Differentiate once with respect to $x$. Treat $y$ as a function of $x$. Differentiating term by term:

   $\frac{2x}{a^2} + \frac{2y}{b^2} \, y' = 0.$

   Divide through by 2:

   $\frac{x}{a^2} + \frac{y}{b^2} \, y' = 0. \qquad(1)$

   3. Differentiate a second time. Differentiate (1) with respect to $x$ (using the product rule on the second term):

   $\frac{1}{a^2} + \frac{1}{b^2} \left( (y')^2 + y \, y'' \right) = 0. \qquad(2)$

   4. Eliminate $a^2$ and $b^2$. From (1), we can write

   $\frac{x}{a^2} = -\frac{y}{b^2} \, y' \quad \Rightarrow \quad \frac{1}{a^2} = -\frac{y}{b^2} \cdot \frac{y'}{x}.$

   Substitute this expression for $\frac{1}{a^2}$ into (2):

   $-\frac{y}{b^2} \cdot \frac{y'}{x} + \frac{1}{b^2} \left( (y')^2 + y y'' \right) = 0.$

   Multiply through by $b^2$ (which is nonzero):

   $$$$

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10. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.If the equation of the curve which passes through the point (1,1) satisfies the differential equation $\dfrac{dy}{dx} = \dfrac{2x-5y+3}{5x+2y-3}$, then the equation of that curve is (A) $x^2 + 5xy - y^2 + 3x - 3y - 5 = 0$ (B) $x^2 + 5xy - y^2 + 3x + 3y - 11 = 0$ (C) $x^2 - 5xy - y^2 - 3x - 3y + 11 = 0$ (D) $x^2 - 5xy - y^2 + 3x + 3y - 1 = 0$

    ›Reveal solutionSolution

    The differential equation is homogeneous after shifting the origin to the intersection of the lines in the numerator and denominator; solving via the substitution $Y = vX$ in the shifted coordinates and using the given point $(1,1)$ yields the curve $x^2 - 5xy - y^2 + 3x + 3y - 1 = 0$, which is option (D).

    ---

    The key idea: when a differential equation has the form $\frac{dy}{dx} = \frac{ax+by+c}{a'x+b'y+c'}$, it is often homogeneous after a translation that eliminates the constant terms. That translation moves the origin to the intersection point of the two lines $ax+by+c=0$ and $a'x+b'y+c'=0$. Once the constants vanish, the equation becomes homogeneous in the new variables, and we can use the substitution $Y = vX$ in the shifted coordinates.

    Here, the numerator is $2x - 5y + 3$ and the denominator is $5x + 2y - 3$. Their intersection is found by solving:

    $\begin{cases} 2x - 5y + 3 = 0 \\ 5x + 2y - 3 = 0 \end{cases}$

    Multiply the first by 2 and the second by 5:

    $4x - 10y + 6 = 0,\quad 25x + 10y - 15 = 0$

    Adding: $29x - 9 = 0 \Rightarrow x = \frac{9}{29}$. Substituting back: $2\left(\frac{9}{29}\right) - 5y + 3 = 0 \Rightarrow \frac{18}{29} + 3 = 5y \Rightarrow \frac{18+87}{29} = 5y \Rightarrow \frac{105}{29} = 5y \Rightarrow y = \frac{21}{29}$.

    So the intersection point is $\left(\frac{9}{29}, \frac{21}{29}\right)$. Shift the origin there by setting:

    $X = x - \frac{9}{29},\quad Y = y - \frac{21}{29}$

    Then $dx = dX$, $dy = dY$, and the differential equation becomes:

    $\frac{dY}{dX} = \frac{2\left(X+\frac{9}{29}\right) - 5\left(Y+\frac{21}{29}\right) + 3}{5\left(X+\frac{9}{29}\right) + 2\left(Y+\frac{21}{29}\right) - 3}$

    Simplify numerator: $2X + \frac{18}{29} - 5Y - \frac{105}{29} + 3 = 2X - 5Y + \left(\frac{18-105}{29} + 3\right) = 2X - 5Y + \left(-\frac{87}{29} + 3\right) = 2X - 5Y + (-3 + 3) = 2X - 5Y$.

    Denominator: $5X + \frac{45}{29} + 2Y + \frac{42}{29} - 3 = 5X + 2Y + \left(\frac{45+42}{29} - 3\right) = 5X + 2Y + \left(\frac{87}{29} - 3\right) = 5X + 2Y + (3 - 3) = 5X + 2Y$.

    Thus the equation reduces to the homogeneous form:

    $\frac{dY}{dX} = \frac{2X - 5Y}{5X + 2Y}$

    Now set $Y = vX$, so $\frac{dY}{dX} = v + X\frac{dv}{dX}$. Substitute:

    $v + X\frac{dv}{dX} = \frac{2X - 5vX}{5X + 2vX} = \frac{2 - 5v}{5 + 2v}$

    Hence:

    $X\frac{dv}{dX} = \frac{2 - 5v}{5 + 2v} - v = \frac{2 - 5v - v(5+2v)}{5+2v} = \frac{2 - 5v - 5v - 2v^2}{5+2v} = \frac{2 - 10v - 2v^2}{5+2v}$

    Factor 2:

    $X\frac{dv}{dX} = \frac{2(1 - 5v - v^2)}{5+2v}$

    Separate variables:

    $\frac{5+2v}{1 - 5v - v^2}\, dv = \frac{2}{X}\, dX$

    Integrate both sides. Notice that the derivative of the denominator $1 - 5v - v^2$ is $-5 - 2v = -(5+2v)$. So:

    $\int \frac{5+2v}{1 - 5v - v^2}\, dv = -\int \frac{-(5+2v)}{1 - 5v - v^2}\, dv = -\log|1 - 5v - v^2| + C$

    Thus:

    $-\log|1 - 5v - v^2| = 2\log|X| + C$

    Multiply by -1:

    $\log|1 - 5v - v^2| = -2\log|X| - C = \log\left(\frac{1}{X^2}\right) + \text{constant}$

    Exponentiate:

    $|1 - 5v - v^2| = \frac{K}{X^2}$

    where $K = e^{-C} > 0$. Remove the absolute value by allowing $K$ to be any nonzero constant:

    $1 - 5v - v^2 = \frac{K}{X^2}$

    Now substitute back $v = Y/X$:

    $1 - 5\frac{Y}{X} - \frac{Y^2}{X^2} = \frac{K}{X^2}$

    Multiply through by $X^2$:

    $X^2 - 5XY - Y^2 = K$

    Recall $X = x - \frac{9}{29}$, $Y = y - \frac{21}{29}$. So:

    $\left(x - \frac{9}{29}\right)^2 - 5\left(x - \frac{9}{29}\right)\left(y - \frac{21}{29}\right) - \left(y - \frac{21}{29}\right)^2 = K$

    We determine $K$ using the given point $(1,1)$. Substitute $x=1$, $y=1$:

    $\left(1 - \frac{9}{29}\right)^2 - 5\left(1 - \frac{9}{29}\right)\left(1 - \frac{21}{29}\right) - \left(1 - \frac{21}{29}\right)^2 = K$

    Compute:

    $1 - \frac{9}{29} = \frac{20}{29},\quad 1 - \frac{21}{29} = \frac{8}{29}$

    So:

    $\left(\frac{20}{29}\right)^2 - 5\cdot\frac{20}{29}\cdot\frac{8}{29} - \left(\frac{8}{29}\right)^2 = \frac{400}{841} - \frac{800}{841} - \frac{64}{841} = \frac{400 - 800 - 64}{841} = \frac{-464}{841}$

    Thus $K = -\frac{464}{841}$.

    Now the equation is:

    $\left(x - \frac{9}{29}\right)^2 - 5\left(x - \frac{9}{29}\right)\left(y - \frac{21}{29}\right) - \left(y - \frac{21}{29}\right)^2 = -\frac{464}{841}$

    Multiply both sides by $841 = 29^2$ to clear denominators:

    $$$$

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11. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
    Q.The differential equation corresponding to the family of curves $y = ae^{2x} + bx^2$, where $a$ and $b$ are parameters is $(x - x^2)\dfrac{d^2y}{dx^2} + y(2 - 4x) =$ (A) $(1 - 2x^2)\dfrac{dy}{dx}$ (B) $(2 - 4x^2)\dfrac{dy}{dx}$ (C) $(1 - 2x)\dfrac{dy}{dx}$ (D) $(2 - 4x)\dfrac{dy}{dx}$

    ›Reveal solutionSolution

    We eliminate the two parameters $a$ and $b$ by differentiating twice and substituting back, obtaining a second‑order linear ODE. The right‑hand side simplifies to $(2-4x)\frac{dy}{dx}$, so the correct option is (D).

    ---

    Concept & Intuition

    When a family of curves contains two arbitrary constants, the corresponding differential equation must be of order 2. The idea is to differentiate the given equation enough times to “free” the parameters, then eliminate them algebraically. Here $y = a e^{2x} + b x^2$ has $a$ and $b$; we differentiate twice, solve for $a$ and $b$ in terms of $y$ and its derivatives, and substitute back. The result will be a relation among $y$, $y'$, and $y''$ — exactly what the question asks.

    ---

    Step‑by‑step derivation

    1. Write the given family and its first two derivatives

    $y = a e^{2x} + b x^2$

    Differentiate:

    $\frac{dy}{dx} = 2a e^{2x} + 2b x$

    Differentiate again:

    $\frac{d^2y}{dx^2} = 4a e^{2x} + 2b$

    2. Eliminate the constant $b$ From $y'' = 4a e^{2x} + 2b$, we can solve for $b$:

    $2b = y'' - 4a e^{2x} \quad\Rightarrow\quad b = \frac{y'' - 4a e^{2x}}{2}$

    But it’s cleaner to eliminate $b$ by combining $y'$ and $y''$. Notice:

    $y' = 2a e^{2x} + 2b x$

    Multiply $y''$ by $x$:

    $x y'' = 4a x e^{2x} + 2b x$

    Subtract $x y''$ from $y'$:

    $y' - x y'' = (2a e^{2x} + 2b x) - (4a x e^{2x} + 2b x) = 2a e^{2x} - 4a x e^{2x} = 2a e^{2x}(1 - 2x)$

    So we have:

    $y' - x y'' = 2a e^{2x}(1 - 2x) \qquad(1)$

    3. Eliminate the constant $a$ From the original equation $y = a e^{2x} + b x^2$, we can also express $a e^{2x}$:

    $a e^{2x} = y - b x^2$

    But we already have $b$ from $y''$: $b = \frac{y'' - 4a e^{2x}}{2}$. Substituting that back would reintroduce $a$. Instead, use equation (1) to solve for $a e^{2x}$:

    $2a e^{2x} = \frac{y' - x y''}{1 - 2x} \quad\Rightarrow\quad a e^{2x} = \frac{y' - x y''}{2(1 - 2x)}$

    Now plug this into the expression for $b$ from $y''$:

    $y'' = 4a e^{2x} + 2b \quad\Rightarrow\quad 2b = y'' - 4a e^{2x} = y'' - 4\cdot\frac{y' - x y''}{2(1 - 2x)}$

    Simplify:

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

    Multiply both sides by $(1 - 2x)$:

    $2b(1 - 2x) = y''(1 - 2x) - 2(y' - x y'')$

    Expand the right side:

    $y''(1 - 2x) - 2y' + 2x y'' = y'' - 2x y'' - 2y' + 2x y'' = y'' - 2y'$

    So:

    $2b(1 - 2x) = y'' - 2y' \quad\Rightarrow\quad b = \frac{y'' - 2y'}{2(1 - 2x)} \qquad(2)$

    4. Substitute $a e^{2x}$ and $b$ back into the original equation The original $y = a e^{2x} + b x^2$ becomes:

    $y = \frac{y' - x y''}{2(1 - 2x)} + \frac{y'' - 2y'}{2(1 - 2x)} \cdot x^2$

    Multiply both sides by $2(1 - 2x)$:

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

    Expand the right side:

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

    So:

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

    5. Rearrange to match the question’s form Bring the $y''$ term to the left: $$$$

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.If A and B are arbitrary constants, then the differential equation having $y = Ae^{-x} + B\cos x$ as its general solution is (A) $(\sin x - \cos x)\dfrac{d^2y}{dx^2} + 2\cos x \dfrac{dy}{dx} - (\sin x + \cos x)y = 0$ (B) $(\cos x - \sin x)\dfrac{d^2y}{dx^2} + 2\cos x \dfrac{dy}{dx} + (\sin x + \cos x)y = 0$ (C) $(\cos x + \sin x)\dfrac{d^2y}{dx^2} + 2\sin x \dfrac{dy}{dx} - (\sin x - \cos x)y = 0$ (D) $(\cos x - \sin x)\dfrac{d^2y}{dx^2} - 2\sin x \dfrac{dy}{dx} + (\cos x + \sin x)y = 0$

    ›Reveal solutionSolution

    The key idea is to eliminate the arbitrary constants $A$ and $B$ from $y = Ae^{-x} + B\cos x$ by differentiating twice and solving the resulting linear system. The correct differential equation is option (B).

    We start with the given general solution:

    $y = A e^{-x} + B \cos x$, where $A$ and $B$ are arbitrary constants.

    To find the differential equation that has this as its general solution, we need to eliminate $A$ and $B$. Since there are two constants, we will need up to the second derivative.

    Concept & Intuition

    A general solution with two arbitrary constants corresponds to a second‑order linear ODE. The constants are “hidden” in the expression; differentiating gives us equations that involve them. By treating $A$ and $B$ as unknowns, we can solve for them from the first two derivatives and substitute back, or directly combine the equations to eliminate them. The trick is to notice that $e^{-x}$ and $\cos x$ are linearly independent, so the elimination will yield a unique linear relation among $y$, $y'$, and $y''$ with coefficients that may depend on $x$.

    ---

    Step‑by‑step elimination

    1. Write down the function and its first two derivatives.

    $y = A e^{-x} + B \cos x$

    Differentiate:

    $y' = -A e^{-x} - B \sin x$

    Differentiate again:

    $y'' = A e^{-x} - B \cos x$

    2. Notice a pattern: we can isolate $A e^{-x}$ and $B \cos x$. From $y$ and $y''$ we have:

    $y = A e^{-x} + B \cos x$

    $y'' = A e^{-x} - B \cos x$

    Adding these two equations gives:

    $y + y'' = 2A e^{-x} \quad \Rightarrow \quad A e^{-x} = \frac{y + y''}{2}$

    Subtracting the second from the first gives:

    $y - y'' = 2B \cos x \quad \Rightarrow \quad B \cos x = \frac{y - y''}{2}$

    3. Now use the first derivative to link these. From $y' = -A e^{-x} - B \sin x$, substitute the expressions we found:

    $y' = -\frac{y + y''}{2} - B \sin x$

    But we also have $B \cos x = \frac{y - y''}{2}$, so $B = \frac{y - y''}{2\cos x}$ (provided $\cos x \neq 0$, but the final ODE will hold everywhere by continuity). Then:

    $y' = -\frac{y + y''}{2} - \frac{y - y''}{2\cos x} \cdot \sin x$

    Simplify:

    $y' = -\frac{y + y''}{2} - \frac{(y - y'')\tan x}{2}$

    4. Clear denominators and rearrange. Multiply through by 2:

    $2y' = -(y + y'') - (y - y'')\tan x$

    $2y' = -y - y'' - y\tan x + y''\tan x$

    Group terms involving $y''$ and $y$:

    $2y' = -y(1 + \tan x) + y''(\tan x - 1)$

    Multiply both sides by $\cos x$ to eliminate the tangent (since $\tan x = \sin x / \cos x$):

    $2y' \cos x = -y(\cos x + \sin x) + y''(\sin x - \cos x)$ …

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