Q.Solve the differential equation $y e^{x/y}\, dx = \left(x e^{x/y} + y^2\right) dy\ (y \neq 0)$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.At a point $P(x,y)$ on a curve $x = f(y)$, the $x$-intercept of the tangent is always equal to the $y$-coordinate of the point of contact, then $f(y) =$ (A) $e^{cy^2}$ (B) $y\log\left(\dfrac{c}{y}\right)$ (C) $cy^2$ (D) $\sin(c + y)$

   ›Reveal solutionSolution

   Translating the given tangent-intercept condition into a differential equation in $x=f(y)$ and solving the resulting linear ODE gives $f(y)=y\log(c/y)$.

   Concept and Intuition

   A geometric condition on the tangent line ("the $x$-intercept equals ...") always converts into an ODE by writing the general tangent line at $(x_0,y_0)$ and reading off where it crosses the axis in question, then equating to the stated condition.

   Step-by-Step Solution

   1. Curve: $x=f(y)$, so $\dfrac{dx}{dy}=f'(y)$, hence slope $\dfrac{dy}{dx}=\dfrac{1}{f'(y)}$.
   2. Tangent at $(x_0,y_0)$: $Y-y_0=\dfrac{1}{f'(y_0)}(X-x_0)$.
   3. $x$-intercept ($Y=0$): $X=x_0-y_0f'(y_0)$.
   4. Given condition: this $x$-intercept equals the $y$-coordinate of the point of contact, i.e. $X=y_0$: $x_0-y_0f'(y_0)=y_0\ \Rightarrow\ f'(y_0)=\dfrac{x_0-y_0}{y_0}=\dfrac{x_0}{y_0}-1$.
   5. Writing as an ODE in $x(y)$: $\dfrac{dx}{dy}-\dfrac{x}{y}=-1$, linear with $P(y)=-\dfrac1y$, $Q(y)=-1$.
   6. Integrating factor $=e^{\int -1/y\,dy}=e^{-\ln y}=\dfrac1y$. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} = \dfrac{2xy - 3y^2}{2x^2 + 3xy}$ is (A) $3\log\left|\dfrac{y}{x}\right| = \dfrac{x}{y} + c$ (B) $\log|xy| = 2xy + c$ (C) $3\log|xy| = \dfrac{2x}{y} + c$ (D) $\log\left|\dfrac{y}{x}\right| = xy + c$

   ›Reveal solutionSolution

   The DE is homogeneous of degree 0; substituting $y=vx$ and separating variables leads, after converting back to $x,y$, to $3\log|xy|=\dfrac{2x}{y}+c$.

   Concept and Intuition

   A first-order DE $\dfrac{dy}{dx}=\dfrac{P(x,y)}{Q(x,y)}$ is homogeneous when $P$ and $Q$ are both homogeneous of the same degree — here both numerator ($2xy-3y^2$) and denominator ($2x^2+3xy$) are degree 2. The standard technique is the substitution $y=vx$, which converts the equation into one in $v$ and $x$ alone that separates.

   Step-by-Step Solution

   1. Given: $\dfrac{dy}{dx} = \dfrac{2xy-3y^2}{2x^2+3xy}$. Both numerator and denominator are homogeneous of degree 2, so substitute $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
   2. Divide numerator and denominator by $x^2$:

   $\frac{2xy-3y^2}{2x^2+3xy} = \frac{2v-3v^2}{2+3v}$

   3. So $v + x\dfrac{dv}{dx} = \dfrac{2v-3v^2}{2+3v}$.
   4. Isolate the derivative term:

   $x\frac{dv}{dx} = \frac{2v-3v^2}{2+3v} - v = \frac{2v-3v^2-v(2+3v)}{2+3v}=\frac{2v-3v^2-2v-3v^2}{2+3v} = \frac{-6v^2}{2+3v}$

   5. Separate variables:

   $\frac{2+3v}{v^2}\,dv = \frac{-6}{x}\,dx \;\Longrightarrow\; \left(\frac{2}{v^2}+\frac{3}{v}\right)dv = -\frac{6}{x}\,dx$

   6. Integrate both sides:

   $-\frac{2}{v}+3\log|v| = -6\log|x| + C$

   7. Rearranging: $-\dfrac2v + 3\log|v| + 6\log|x| = C$. Since $6\log|x|=3\log(x^2)$, combine logs: …

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3. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The general solution of the differential equation $(x+y-1)\,dy = (x-y+1)\,dx$ is (A) $x^2 - 2xy - y^2 + 2x + 2y + c = 0$ (B) $x^2 + 2xy - y^2 + 2x + 2y + c = 0$ (C) $x^2 + 2xy + y^2 + 2x + 2y + c = 0$ (D) $x^2 - 2xy - y^2 + 2x - 2y + c = 0$

   ›Reveal solutionSolution

   Shifting the origin to remove the constant terms turns this into a homogeneous DE in $X,Y$; solving it and substituting back gives $x^2-2xy-y^2+2x+2y+c=0$.

   Concept and Intuition

   A DE of the form $\dfrac{dy}{dx}=\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$ (with $a_1b_2\ne a_2b_1$) is reduced to a homogeneous equation by shifting the origin to the intersection point of the two lines $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$. This removes the constant terms, and the resulting equation in the shifted variables is exactly homogeneous and solvable by $Y=vX$.

   Step-by-Step Solution

   1. Given $(x+y-1)dy=(x-y+1)dx$, i.e. $\dfrac{dy}{dx}=\dfrac{x-y+1}{x+y-1}$.
   2. Shift $x=X+h,\ y=Y+k$ to kill the constants. We need:

   $h-k+1=0,\qquad h+k-1=0$

   Adding: $2h=0\Rightarrow h=0$; then $k=1$.

   3. With $x=X,\ y=Y+1$: numerator $x-y+1 = X-Y$, denominator $x+y-1=X+Y$. So:

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

   This is homogeneous of degree 0.

   4. Let $Y=vX$, $\dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$:

   $v+X\frac{dv}{dX} = \frac{1-v}{1+v}$

   5. Isolate: $X\dfrac{dv}{dX} = \dfrac{1-v}{1+v}-v = \dfrac{1-v-v-v^2}{1+v}=\dfrac{1-2v-v^2}{1+v}$.
   6. Separate: $\dfrac{1+v}{1-2v-v^2}\,dv = \dfrac{dX}{X}$. Let $w=1-2v-v^2$, $dw=-2(1+v)\,dv$, so $(1+v)dv=-\tfrac12 dw$:

   $-\frac12\int\frac{dw}{w} = \int\frac{dX}{X} \;\Longrightarrow\; -\frac12\log|w| = \log|X|+C_1$

   7. So $\log|w| = -2\log|X|+C_2 \Rightarrow w\,X^2 = C_3$, i.e. $(1-2v-v^2)X^2=C_3$. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.The general solution of the differential equation $x\left(\dfrac{dy}{dx}\right)^2 + 2\sqrt{xy}\dfrac{dy}{dx} + y = 0$ is (A) $x+y=c$ (B) $\sqrt{x}+\sqrt{y}=\sqrt{c}$ (C) $x^2+y^2=c^2$ (D) $\sqrt{x}-\sqrt{y}=\sqrt{c}$

   ›Reveal solutionSolution

   The equation is a perfect-square quadratic in $dy/dx$ (discriminant zero), which reduces it to a simple separable equation whose solution is $\sqrt x+\sqrt y=\sqrt c$.

   Concept and Intuition

   When a differential equation is quadratic in $p=dy/dx$, check the discriminant. If it is a perfect square (or zero), the quadratic factors nicely, collapsing the equation to a single, much simpler first-order relation that can be solved by separation of variables.

   Step-by-Step Solution

   1. Given: $x\left(\dfrac{dy}{dx}\right)^2 + 2\sqrt{xy}\dfrac{dy}{dx} + y = 0$. Let $p=dy/dx$: $xp^2+2\sqrt{xy}\,p+y=0$.
   2. Treat as a quadratic in $p$: discriminant $= (2\sqrt{xy})^2 - 4(x)(y) = 4xy - 4xy = 0$.
   3. Since the discriminant is zero, $p = \dfrac{-2\sqrt{xy}}{2x} = -\dfrac{\sqrt{xy}}{x} = -\sqrt{\dfrac{y}{x}}$ (using $\sqrt{xy}/x = \sqrt{y}/\sqrt{x}$).
   4. So $\dfrac{dy}{dx} = -\sqrt{\dfrac{y}{x}}$, which separates as $\dfrac{dy}{\sqrt y} = -\dfrac{dx}{\sqrt x}$.
   5. Integrate both sides: $2\sqrt y = -2\sqrt x + C_1 \Rightarrow \sqrt x + \sqrt y = \dfrac{C_1}{2}$. …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.The general solution of the differential equation $y(2x + y)\, dx = x(x + y)\, dy$ is (A) $\log\left(\dfrac{cx^2}{y}\right) + \dfrac{y}{x} = 0$ (B) $\log\left(\dfrac{y}{cx^2}\right) + \dfrac{y}{x} = 0$ (C) $\log(cx^2y^2) + \dfrac{y}{x} = 0$ (D) $\log(cx^2y) + \dfrac{y}{x} = 0$

   ›Reveal solutionSolution

   A homogeneous ODE solved by the substitution $y=vx$; simplifying and integrating gives $\log\left(\dfrac{y}{cx^2}\right)+\dfrac{y}{x}=0$.

   Concept and Intuition

   $y(2x+y)dx=x(x+y)dy$ is homogeneous of degree 2 on both sides, so the standard substitution $y=vx$ (with $dy=v\,dx+x\,dv$) reduces it to a separable equation in $v$ and $x$.

   Step-by-Step Solution

   1. Put $y=vx$, $dy=v\,dx+x\,dv$.
   2. LHS: $y(2x+y)=vx(2x+vx)=vx^2(2+v)$.
   3. RHS: $x(x+y)\,dy=x\cdot x(1+v)(v\,dx+x\,dv)=x^2(1+v)(v\,dx+x\,dv)$.
   4. Equation becomes $vx^2(2+v)\,dx=x^2(1+v)(v\,dx+x\,dv)$. Divide by $x^2$: $v(2+v)dx=v(1+v)dx+x(1+v)dv$.
   5. Subtract $v(1+v)dx$ from both sides: $v[(2+v)-(1+v)]dx=x(1+v)dv \Rightarrow v\,dx=x(1+v)\,dv$.
   6. Separate: $\dfrac{dx}{x}=\dfrac{1+v}{v}dv=\left(\dfrac1v+1\right)dv$.
   7. Integrate: $\log x=\log v+v+C \Rightarrow \log\dfrac{x}{v}-v=C$.
   8. Substitute back $v=y/x$, so $x/v=x^2/y$: $\log\dfrac{x^2}{y}-\dfrac{y}{x}=C$. …

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6. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.The general solution of the differential equation $\left(x\sin\dfrac{y}{x}\right)dy=\left(y\sin\dfrac{y}{x}-x\right)dx$ is (A) $\log x+\tan\dfrac{y}{x}=c$ (B) $\log x+\cos\dfrac{y}{x}=c$ (C) $\log x-\sin\dfrac{y}{x}=c$ (D) $\log x-\cos\dfrac{y}{x}=c$

   ›Reveal solutionSolution

   This is a homogeneous first-order ODE solved via $y=vx$; the answer is (D).

   Concept and Intuition

   When every term in the ODE is a function of $y/x$ times some power of $x$ or $y$ (a homogeneous equation), the substitution $y=vx$ (so $v=y/x$) reduces it to a separable equation in $v$ and $x$.

   Step-by-Step Solution

   1. The given equation is $\left(x\sin\frac yx\right)dy=\left(y\sin\frac yx-x\right)dx$, so

   $\frac{dy}{dx}=\frac{y\sin(y/x)-x}{x\sin(y/x)}.$

   2. Substitute $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. The right side becomes $\dfrac{vx\sin v - x}{x\sin v}=\dfrac{v\sin v-1}{\sin v}=v-\dfrac1{\sin v}$.
   3. So $v+x\dfrac{dv}{dx}=v-\dfrac1{\sin v}$, giving $x\dfrac{dv}{dx}=-\dfrac1{\sin v}$.
   4. Separate variables: $\sin v\,dv=-\dfrac{dx}{x}$. …

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7. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{x+y+1}{x-3y+5} = 0$ is (A) $3(y-1)^2 - 2(x+2)(y-1) - (x+2)^2 = c$ (B) $x^2 - 3y^2 - 4xy - 2x - 10y = c$ (C) $3(y+1)^2 + 2(x-2)(y+1) - (x-2)^2 = c$ (D) $x^2 + 3y^2 + 4xy + 2x + 10y = c$

   ›Reveal solutionSolution

   This is a differential equation of the form $\dfrac{dy}{dx}=-\dfrac{ax+by+c}{a'x+b'y+c'}$ with non-parallel linear terms; shifting the origin to the intersection of the two lines reduces it to a homogeneous equation. The answer is (A).

   Concept and Intuition

   When a differential equation has the form $\dfrac{dy}{dx}=-\dfrac{x+y+1}{x-3y+5}$, the numerator and denominator are linear in $x,y$ but not proportional, so it isn't directly homogeneous. The standard trick is to translate the axes to the point where the two lines $x+y+1=0$ and $x-3y+5=0$ intersect — in the new coordinates the equation becomes exactly homogeneous of degree one, which we can solve with the substitution $Y=vX$.

   Step-by-Step Solution

   1. Find the intersection point. Solve $x+y=-1$ and $x-3y=-5$ simultaneously. Subtracting: $4y=4\Rightarrow y=1$, then $x=-2$. So the lines meet at $(-2,1)$.
   2. Shift coordinates: let $X=x+2,\ Y=y-1$ (so $dX=dx,\ dY=dy$). The equation becomes $\dfrac{dY}{dX}=-\dfrac{X+Y}{X-3Y}$, which is homogeneous (every term is degree 1 in $X,Y$).
   3. Substitute $Y=vX$, so $\dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$. Then

   $v+X\frac{dv}{dX}=-\frac{1+v}{1-3v}\ \Rightarrow\ X\frac{dv}{dX}=\frac{3v^2-2v-1}{1-3v}.$

   4. Separate variables:

   $\frac{1-3v}{3v^2-2v-1}\,dv=\frac{dX}{X}.$

   Factor the denominator: $3v^2-2v-1=(3v+1)(v-1)$. Partial fractions give

   $\frac{1-3v}{(3v+1)(v-1)}=\frac{-3/2}{3v+1}+\frac{-1/2}{v-1}.$

   5. Integrate both sides:

   $-\tfrac12\log|3v+1|-\tfrac12\log|v-1|=\log X + C_1$

   $\Rightarrow\ -2\log X=\log|3v+1|+\log|v-1|+C_2\ \Rightarrow\ X^{-2}=K(3v+1)(v-1).$ …

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8. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} = \dfrac{x+y}{x-y}$ is (A) $y - x = cx^2$ (B) $\mathrm{Tan}^{-1}\left(\dfrac{y}{x}\right)=\log\left(cx\sqrt{x^2+y^2}\right)$ (C) $x+y=cx^2$ (D) $\mathrm{Tan}^{-1}\left(\dfrac{y}{x}\right)=\log\left(c\sqrt{x^2+y^2}\right)$

   ›Reveal solutionSolution

   The equation is homogeneous of degree $0$; the substitution $y=vx$ separates variables and integrates to the polar-flavoured solution $\mathrm{Tan}^{-1}(y/x)=\log\left(c\sqrt{x^2+y^2}\right)$.

   Concept and Intuition

   $\dfrac{dy}{dx}=\dfrac{x+y}{x-y}$ is homogeneous (both numerator and denominator scale the same way under $x\to\lambda x,y\to\lambda y$), so $y=vx$ turns it into a separable equation in $v$ and $x$. The appearance of both $\mathrm{Tan}^{-1}(y/x)$ and $\log(x^2+y^2)$ afterwards is typical whenever the separated integral produces both an arctangent and a logarithm term.

   Step-by-Step Solution

   1. Let $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Then $v+x\dfrac{dv}{dx}=\dfrac{1+v}{1-v}$.
   2. $x\dfrac{dv}{dx}=\dfrac{1+v}{1-v}-v=\dfrac{1+v-v(1-v)}{1-v}=\dfrac{1+v^2}{1-v}$.
   3. Separate: $\dfrac{1-v}{1+v^2}\,dv=\dfrac{dx}{x}$, i.e. $\left(\dfrac1{1+v^2}-\dfrac{v}{1+v^2}\right)dv=\dfrac{dx}{x}$.
   4. Integrate: $\mathrm{Tan}^{-1}v-\dfrac12\log(1+v^2)=\log x + C$.
   5. So $\mathrm{Tan}^{-1}v=\log x+\dfrac12\log(1+v^2)+C=\log\!\left(x\sqrt{1+v^2}\right)+C$. …

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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.The general solution of the differential equation $\frac{dy}{dx} = \frac{2x^2 - xy - y^2}{x^2 - y^2}$ is (A) $\log\left|\frac{y^2-2x^2}{x^2}\right| + \sqrt{2}\log\left|\frac{y-\sqrt{2}x}{y+\sqrt{2}x}\right| + 2\sqrt{2}\log|x| = c$ (B) $\sqrt{2}\log\left|\frac{y^2-2x^2}{x^2}\right| + \log\left|\frac{y-\sqrt{2}x}{y+\sqrt{2}x}\right| + 2\sqrt{2}\log|x| = c$ (C) $\sqrt{2}\log\left|\frac{y^2+2x^2}{x^2}\right| + \log\left|\frac{y+\sqrt{2}x}{y-\sqrt{2}x}\right| + 2\sqrt{2}\log|x| = c$ (D) $\log\left|\frac{2x^2-y^2}{x^2}\right| + \sqrt{2}\log\left|\frac{y+\sqrt{2}x}{y-\sqrt{2}x}\right| + \log|x| = c$

   ›Reveal solutionSolution

   A homogeneous ODE solved by $y=vx$ substitution; after partial fractions, the solution is option (B).

   Concept and Intuition

   The right side $\dfrac{2x^2-xy-y^2}{x^2-y^2}$ is a ratio of degree-2 homogeneous polynomials in $x,y$, so dividing top and bottom by $x^2$ turns everything into a function of $v=y/x$ alone — the standard homogeneous-equation substitution $y=vx$.

   Step-by-Step Solution

   1. Let $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Dividing numerator and denominator of the RHS by $x^2$:

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

   2. Isolate $x\,dv/dx$:

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

   3. Factor the cubic: $v^3-v^2-2v+2=(v-1)(v^2-2)$, and $1-v^2=-(v-1)(v+1)$. The $(v-1)$ cancels:

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

   4. Separate variables: $\dfrac{v+1}{2-v^2}dv=\dfrac{dx}{x}$.
   5. Partial fractions on $2-v^2=(\sqrt2-v)(\sqrt2+v)$: writing $\dfrac{v+1}{(\sqrt2-v)(\sqrt2+v)}=\dfrac{A}{\sqrt2-v}+\dfrac{B}{\sqrt2+v}$ gives $A=\dfrac{2+\sqrt2}{4}$, $B=\dfrac{\sqrt2-2}{4}$.
   6. Integrating and simplifying logs (combining $\log|\sqrt2-v|\pm\log|\sqrt2+v|$ into $\log|2-v^2|$ and $\log\left|\frac{\sqrt2-v}{\sqrt2+v}\right|$ terms) and multiplying through, one obtains: $\sqrt2\log|2-v^2| + \log\left|\frac{\sqrt2-v}{\sqrt2+v}\right| + 2\sqrt2\log|x| = C.$ …

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10. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx}=\dfrac{2xy-4x+y-2}{2xy+x-4y-2}$ is (A) $5(y-x)+2\log\left(\dfrac{y-2}{x-2}\right)=c$ (B) $2(y-x)-5\log\left(\dfrac{y-2}{x-2}\right)=c$ (C) $2(y-x)+5\log\left(\dfrac{y-2}{x-2}\right)=c$ (D) $5(y-x)-2\log\left(\dfrac{y-2}{x-2}\right)=c$

    ›Reveal solutionSolution

    This tests solving a separable ODE after factoring both the numerator and denominator by grouping. Answer: (C).

    Concept and Intuition

    A rational-function differential equation like this often hides a separable form once you notice both numerator and denominator can be grouped into two linear-in-$x$ and linear-in-$y$ factors. Spotting the factoring is the whole trick.

    Step-by-Step Solution

    1. Group the numerator: $2xy-4x+y-2 = 2x(y-2)+1\cdot(y-2) = (y-2)(2x+1)$.
    2. Group the denominator: $2xy+x-4y-2 = x(2y+1)-2(2y+1) = (2y+1)(x-2)$.
    3. So $\dfrac{dy}{dx}=\dfrac{(y-2)(2x+1)}{(x-2)(2y+1)}$, which is separable:

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

    4. Rewrite each side to make integration easy:

    $\frac{2y+1}{y-2} = \frac{2(y-2)+5}{y-2} = 2+\frac{5}{y-2}, \qquad \frac{2x+1}{x-2} = 2+\frac{5}{x-2}.$

    5. Integrate both sides: $2y + 5\log|y-2| = 2x + 5\log|x-2| + C.$ …

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11. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.The general solution of the differential equation $(x - (x+y)\log(x+y))dx + x\,dy = 0$ is (A) $y\log(x+y) = cx$ (B) $x\log(x+y) = cy$ (C) $\log(x+y) = cy$ (D) $\log(x+y) = cx$

    ›Reveal solutionSolution

    This tests recognizing that regrouping terms via $v=x+y$ converts an awkward mixed equation into a separable one. The general solution is $\log(x+y)=cx$, option (D).

    Concept and Intuition

    The given equation mixes $x$, $y$, and $\log(x+y)$ in a way that isn't obviously separable or linear at first glance. The key insight is spotting that $dx+dy$ (which appears once you regroup) is exactly $d(x+y)$ — so substituting $v=x+y$ collapses the two-variable equation into one purely in $v$ and $x$, which then separates cleanly.

    Step-by-Step Solution

    1. Expand the given equation: $x\,dx-(x+y)\log(x+y)\,dx+x\,dy=0$.
    2. Regroup: $x(dx+dy)=(x+y)\log(x+y)\,dx$.
    3. Let $v=x+y$, so $dv=dx+dy$. The equation becomes $x\,dv=v\log v\,dx$.
    4. Separate variables: $\dfrac{dv}{v\log v}=\dfrac{dx}{x}$.
    5. For the left side, let $w=\log v$, $dw=\dfrac{dv}{v}$, so $\dfrac{dv}{v\log v}=\dfrac{dw}{w}$, integrating to $\log|w|=\log|\log v|$.
    6. Integrate both sides: $\log|\log v|=\log|x|+C_1 \Rightarrow \log v = kx$ for some constant $k$ (absorbing $e^{C_1}$ and sign). …

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12. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
    Q.The general solution of the differential equation $\dfrac{dy}{dx} = \dfrac{2x+y-3}{2y-x+3}$ is (A) $x^2 - xy - y^2 + 3x + 3y + c = 0$ (B) $x^2 - xy - y^2 - 3x - 3y + c = 0$ (C) $x^2 + xy - y^2 - 3x - 3y + c = 0$ (D) $x^2 + xy + y^2 + 3x - 3y + c = 0$

    ›Reveal solutionSolution

    The ODE rearranges into an exact differential equation; solving it directly by the exact-equation method gives $x^2+xy-y^2-3x-3y+c=0$.

    Concept and Intuition

    $\dfrac{dy}{dx}=\dfrac{2x+y-3}{2y-x+3}$ is linear in $x$ and $y$ in both numerator and denominator, which is a strong hint to try the exact differential equation test: cross-multiply into $M\,dx+N\,dy=0$ form and check if $\partial M/\partial y=\partial N/\partial x$.

    Step-by-Step Solution

    1. Cross-multiplying: $(2y-x+3)\,dy=(2x+y-3)\,dx \Rightarrow (2x+y-3)\,dx-(2y-x+3)\,dy=0$.
    2. Rewrite with a plus sign: $(2x+y-3)\,dx+(x-2y-3)\,dy=0$, so $M=2x+y-3$ and $N=x-2y-3$.
    3. Check exactness: $\dfrac{\partial M}{\partial y}=1$ and $\dfrac{\partial N}{\partial x}=1$ — equal, so the equation is exact.
    4. Find $F(x,y)$ with $F_x=M$: integrate w.r.t. $x$ (treating $y$ as constant):

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

    5. Differentiate w.r.t. $y$ and match to $N$: $F_y=x+g'(y)=x-2y-3 \Rightarrow g'(y)=-2y-3$.
    6. Integrate: $g(y)=-y^2-3y$ (constant absorbed later).
    7. So $F(x,y)=x^2+xy-3x-y^2-3y$, and the general solution is $F=c$:

    $x^2+xy-y^2-3x-3y=c \iff x^2+xy-y^2-3x-3y+c=0.$

    Common Mistakes …

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