Q.Solve the following differential equation: $(x^2 + xy) dy = (x^2 + y^2) dx$

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.The general solution of the differential equation $(xy+y^2)dx-(x^2-2xy)dy=0$ is (A) $cxy^2=e^{x/y}$ (B) $cxy^2e^{x/y}=1$ (C) $cxye^{x/y}=1$ (D) $cxy=e^{x/y}$

   ›Reveal solutionSolution

   This is a homogeneous first-order ODE; the substitution $y=vx$ solves it, giving $cxy^2e^{x/y}=1$, option (B).

   Concept and Intuition

   Both $(xy+y^2)$ and $(x^2-2xy)$ are homogeneous of degree 2 in $x,y$, so the standard substitution $y=vx$ (turning the equation into one in $v$ and $x$ alone) separates variables.

   Step-by-Step Solution

   1. Write the ODE as $(xy+y^2)\,dx=(x^2-2xy)\,dy$. Substitute $y=vx$, $dy=v\,dx+x\,dv$.
   2. $xy+y^2=x^2(v+v^2)$ and $x^2-2xy=x^2(1-2v)$. The equation becomes

   $x^2(v+v^2)\,dx=x^2(1-2v)(v\,dx+x\,dv).$

   3. Divide by $x^2$ and collect $dx$ terms: $\big[(v+v^2)-v(1-2v)\big]dx=(1-2v)x\,dv$. The bracket simplifies to $3v^2$, so

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

   4. Integrate: $\displaystyle\log x=\frac13\int\left(\frac{1}{v^2}-\frac2v\right)dv=\frac13\left(-\frac1v-2\log v\right)+C$.
   5. Multiply by 3: $3\log x+2\log v+\dfrac1v=C'$, i.e. $\log(x^3v^2)=C'-\dfrac1v$, so $x^3v^2=Ke^{-1/v}$.
   6. Substitute back $v=y/x$: $x^3\cdot\dfrac{y^2}{x^2}=xy^2$ and $1/v=x/y$, giving $xy^2=Ke^{-x/y}$, i.e. $cxy^2e^{x/y}=1$ (absorbing constants).

   Common Mistakes

   • Sign error in simplifying $(v+v^2)-v(1-2v)$ (it's $+3v^2$, not $v^2$ or $-3v^2$).
   • Forgetting to convert $v$ back to $y/x$ correctly in the exponential term ($1/v=x/y$, easy to invert incorrectly).

   ✓Final answer

   The correct option is (B) — $cxy^2e^{x/y}=1$.

   ANSWER: B

2. 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 $x/v=x^2/y$: $\log\dfrac{x^2}{y}-\dfrac{y}{x}=C$.
   9. Since $C$ is an arbitrary constant, we may equally write $-C=\log c$: negating the whole equation, $\log\dfrac{y}{x^2}+\dfrac{y}{x}=-C=\log c$, i.e. $\log\dfrac{y}{cx^2}+\dfrac{y}{x}=0$, matching the printed answer form.

   Common Mistakes

   • Sign errors when combining $v(2+v)dx$ and $v(1+v)dx$ — the whole simplification collapses to the clean $v\,dx=x(1+v)dv$ only if signs are tracked carefully.
   • Not recognising that flipping the sign of the arbitrary constant of integration is legitimate (since $c$ is arbitrary), which is needed to match the answer's printed form.

   ✓Final answer

   The correct option is (B) — $\log\left(\dfrac{y}{cx^2}\right)+\dfrac{y}{x}=0$.

   ANSWER: B

3. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.The general solution of the differential equation $x\,dy - y\,dx = \sqrt{x^2 + y^2}\,dx$ is (A) $y + \sqrt{x^2 + y^2} = cx^2$ (B) $y + \sqrt{x^2 + y^2} = cx$ (C) $x + \sqrt{x^2 + y^2} = cy$ (D) $x - \sqrt{x^2 + y^2} = cy^2$

   ›Reveal solutionSolution

   A homogeneous differential equation, solved with the substitution $y=vx$. Answer: $y+\sqrt{x^2+y^2}=cx^2$.

   Concept and Intuition

   Every term here — $x\,dy$, $y\,dx$, and $\sqrt{x^2+y^2}\,dx$ — scales the same way if $x,y$ are both scaled by the same factor, which is the signature of a homogeneous equation. The standard substitution $y=vx$ converts it into a separable equation in $v$ and $x$.

   Step-by-Step Solution

   1. Write $\dfrac{dy}{dx} = \dfrac{y+\sqrt{x^2+y^2}}{x}$ (dividing through by $dx$, taking $x>0$).
   2. Let $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. Then: $v+x\frac{dv}{dx} = \frac{vx+\sqrt{x^2+v^2x^2}}{x} = v+\sqrt{1+v^2}.$
   3. So $x\dfrac{dv}{dx}=\sqrt{1+v^2}$, which separates: $\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}$.
   4. Integrate: $\log\left(v+\sqrt{1+v^2}\right) = \log x + \log C \Rightarrow v+\sqrt{1+v^2}=Cx$.
   5. Substitute back $v=y/x$: $\dfrac{y}{x}+\sqrt{1+\dfrac{y^2}{x^2}} = Cx \Rightarrow \dfrac{y+\sqrt{x^2+y^2}}{x}=Cx \Rightarrow y+\sqrt{x^2+y^2}=Cx^2$.

   Common Mistakes

   • Forgetting to multiply through by $x$ at the final step, mistakenly leaving the constant with only one power of $x$ (option B).
   • Sign error in identifying $\int \dfrac{dv}{\sqrt{1+v^2}} = \log(v+\sqrt{1+v^2})$, a standard but easy-to-misremember integral.

   ✓Final answer

   The correct option is (A) — $y + \sqrt{x^2 + y^2} = cx^2$.

   ANSWER: A

4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.The general solution of the differential equation $(x+y)y\,dx+(y-x)x\,dy=0$ is (A) $x+y\log(cy)=0$ (B) $\dfrac{y}{x}=\log(xy)+c$ (C) $x+y\log(cxy)=0$ (D) $\dfrac{y}{x}=\log(cxy)$

   ›Reveal solutionSolution

   This is a homogeneous first-order ODE; the substitution $y=vx$ turns it into a separable equation whose integral rearranges to $x+y\log(cxy)=0$.

   Concept and Intuition

   Every term of $(x+y)y\,dx+(y-x)x\,dy=0$ is of the same total degree (degree 2 in $x,y$), which is the signature of a homogeneous differential equation. The standard technique is to set $y=vx$, turning the equation into one involving only $v$ and $x$, which separates.

   Step-by-Step Solution

   1. Expand: $(xy+y^2)\,dx+(xy-x^2)\,dy=0$ — every term has degree 2, confirming homogeneity.
   2. Put $y=vx$, so $dy=v\,dx+x\,dv$. Then $xy+y^2=x^2v(1+v)$ and $xy-x^2=x^2(v-1)$.
   3. Substituting: $x^2v(1+v)\,dx+x^2(v-1)(v\,dx+x\,dv)=0$. Dividing by $x^2$: $v(1+v)\,dx+(v-1)v\,dx+x(v-1)\,dv=0$.
   4. Collect the $dx$ coefficient: $v(1+v)+v(v-1)=v[(1+v)+(v-1)]=2v^2$. So $2v^2\,dx+x(v-1)\,dv=0$.
   5. Separate variables: $\dfrac{dx}{x}=\dfrac{(1-v)}{2v^2}\,dv=\left(\dfrac{1}{2v^2}-\dfrac{1}{2v}\right)dv$.
   6. Integrate: $\log x = -\dfrac{1}{2v}-\dfrac12\log v+C$. Multiply by 2: $2\log x+\log v=-\dfrac1v+2C\Rightarrow \log(x^2v)=-\dfrac1v+K$.
   7. Since $x^2v=x^2\cdot\dfrac{y}{x}=xy$ and $\dfrac1v=\dfrac{x}{y}$: $\log(xy)+\dfrac{x}{y}=K$.
   8. Multiply through by $y$: $y\log(xy)+x=Ky$, i.e. $x+y\big[\log(xy)-K\big]=0$. Writing $\log(xy)-K=\log(xy)-\log(e^K)=\log\!\big(xy\cdot e^{-K}\big)=\log(cxy)$ (with $c=e^{-K}$), this is $x+y\log(cxy)=0$.

   Common Mistakes

   • Forgetting to convert $\dfrac1v$ back to $\dfrac{x}{y}$ and $v$ back to $\dfrac{y}{x}$ before comparing to the answer choices.
   • Losing track of which constant absorbs the $e^{-K}$ term when rewriting $\log(xy)-K$ as $\log(cxy)$.

   ✓Final answer

   The correct option is (C) — $x+y\log(cxy)=0$.

   ANSWER: C

5. 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:

   $\sqrt2\log|2-v^2| + \log\left|\frac{\sqrt2-v}{\sqrt2+v}\right| + 2\sqrt2\log|x| = C.$

   7. Substitute back $v=y/x$: $2-v^2=\dfrac{2x^2-y^2}{x^2}$ and $\dfrac{\sqrt2-v}{\sqrt2+v}=\dfrac{y-\sqrt2x}{y+\sqrt2x}$ (up to sign, absorbed by the modulus), giving exactly option (B).

   Common Mistakes

   • Losing the $(v-1)$ common factor and trying to integrate the un-simplified cubic-over-quadratic directly.
   • Sign slips when converting $\log|\sqrt2-v|-\log|\sqrt2+v|$ into the ratio form, which flips which option (A)/(B)/(C)/(D) the coefficients land on.

   ✓Final answer

   The correct option is (B) — $\sqrt{2}\log\left|\frac{y^2-2x^2}{x^2}\right| + \log\left|\frac{y-\sqrt{2}x}{y+\sqrt{2}x}\right| + 2\sqrt{2}\log|x| = c$.

   ANSWER: B

6. 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: $2\sqrt y = -2\sqrt x + C_1 \Rightarrow \sqrt x + \sqrt y = \dfrac{C_1}{2}$.
   6. Renaming the constant as $\sqrt c$ (a nonnegative constant, consistent with $\sqrt x+\sqrt y\ge0$), the general solution is $\sqrt x + \sqrt y = \sqrt c$.

   Common Mistakes

   • Trying to solve the quadratic in $p$ using the full quadratic formula without noticing the discriminant vanishes, missing the simplification.
   • Sign errors when simplifying $\sqrt{xy}/x$ to $\sqrt{y/x}$.

   ✓Final answer

   The correct option is (B) — $\sqrt{x}+\sqrt{y}=\sqrt{c}$.

   ANSWER: B

7. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.The general solution of xdy - ydx = ydy is (A) $y = Ae^{-x/y}$ (B) $y = Ae^{x}$ (C) $\dfrac{y}{x} = Ae^{x}$ (D) $\dfrac{x}{y} + \dfrac{y}{x} = C$

   ›Reveal solutionSolution

   The ODE is homogeneous; solving with $y=Vx$ and simplifying the resulting logarithmic relation gives $y=Ae^{-x/y}$.

   Concept and Intuition

   "$x\,dy-y\,dx$" style equations are homogeneous in $x,y$ and standardly solved by substituting $y=Vx$, which turns them into separable equations in $V$ and $x$.

   Step-by-Step Solution

   1. Rearrange: $x\,dy-y\,dy=y\,dx \Rightarrow dy(x-y)=y\,dx \Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x-y}$.
   2. Let $y=Vx$, so $\dfrac{dy}{dx}=V+x\dfrac{dV}{dx}$. RHS becomes $\dfrac{Vx}{x-Vx}=\dfrac{V}{1-V}$.
   3. So $V+x\dfrac{dV}{dx}=\dfrac{V}{1-V} \Rightarrow x\dfrac{dV}{dx} = \dfrac{V-V(1-V)}{1-V}=\dfrac{V^2}{1-V}$.
   4. Separate: $\dfrac{1-V}{V^2}dV = \dfrac{dx}{x} \Rightarrow \left(\dfrac1{V^2}-\dfrac1V\right)dV=\dfrac{dx}{x}$.
   5. Integrate: $-\dfrac1V-\log V = \log x+C_1$.
   6. Substitute back $V=y/x$: $-\dfrac{x}{y}-\log\dfrac{y}{x}=\log x+C_1 \Rightarrow -\dfrac{x}{y}-\log y+\log x=\log x+C_1 \Rightarrow -\dfrac{x}{y}-\log y=C_1$.
   7. So $\dfrac{x}{y}=-\log y-C_1 = \log A-\log y=\log\dfrac{A}{y}$ (writing $-C_1=\log A$). Exponentiating: $e^{x/y}=\dfrac{A}{y}\Rightarrow y=Ae^{-x/y}$.

   Common Mistakes

   • Sign slips when substituting $V=y/x$ back into the logarithm — the $\log(y/x)$ term splits into $\log y-\log x$, and the $\log x$ pieces cancel, easy to mishandle.
   • Trying to solve for $y$ explicitly (it's implicit — the answer is naturally in the exponential-implicit form given).

   ✓Final answer

   The correct option is (A) — $y = Ae^{-x/y}$.

   ANSWER: A

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 $\mathrm{Tan}^{-1}v=\log x+\dfrac12\log(1+v^2)+C=\log\!\left(x\sqrt{1+v^2}\right)+C$.
   6. Since $v=y/x$: $x\sqrt{1+v^2}=x\sqrt{1+y^2/x^2}=\sqrt{x^2+y^2}$. So $\mathrm{Tan}^{-1}(y/x)=\log\!\left(c\sqrt{x^2+y^2}\right)$ with $c=e^C$.

   Common Mistakes

   • Losing the extra factor of $x$ when converting $\sqrt{1+v^2}$ back — it correctly cancels to leave just $\sqrt{x^2+y^2}$, no extra $x$ (ruling out the option with $cx\sqrt{x^2+y^2}$).
   • Sign error on the $\dfrac{v}{1+v^2}$ integral (it integrates to $-\tfrac12\log(1+v^2)$, i.e. it subtracts).

   ✓Final answer

   The correct option is (D) — $\mathrm{Tan}^{-1}\left(\dfrac yx\right)=\log\left(c\sqrt{x^2+y^2}\right)$.

   ANSWER: D

9. 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:

   $-\frac2v + 3\log|v x^2| = C$

   8. Substitute back $v=y/x$, so $vx^2=xy$ and $\dfrac1v=\dfrac xy$:

   $3\log|xy| - \frac{2x}{y} = C \;\Longrightarrow\; 3\log|xy| = \frac{2x}{y}+c$

   Common Mistakes

   • Forgetting to convert $6\log|x|$ into $3\log(x^2)$ before combining with $3\log|v|$ — this is essential to get the clean $3\log|xy|$ form that matches the given options.
   • Sign slip when moving $-2/v$ across the equation, which flips the final answer to option (A)'s form instead of (C)'s.

   ✓Final answer

   The correct option is (C) — $3\log|xy| = \dfrac{2x}{y}+c$.

   ANSWER: C

10. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
    Q.The general solution of the differential equation $(x-y-1)\,dy=(x+y+1)\,dx$ is (A) $\tan^{-1}\left(\dfrac{y+1}{x}\right)-\dfrac{1}{2}\log(x^2+y^2+2y+1)=c$ (B) $(x-y)+\log(x+y)=c$ (C) $y^2-x^2+xy-3y-x=c$ (D) $(x-y-1)^2(x+y+1)^3=c$

    ›Reveal solutionSolution

    A shift of origin to where $x+y+1=0$ meets $x-y-1=0$ turns this into a standard homogeneous equation, giving $\tan^{-1}\!\left(\frac{y+1}{x}\right)-\frac12\log(x^2+y^2+2y+1)=c$.

    Concept and Intuition

    Equations of the form $\dfrac{dy}{dx}=\dfrac{ax+by+c}{a'x+b'y+c'}$ that are not homogeneous (because of the constant terms) can be made homogeneous by shifting the origin to the point where the two linear expressions both vanish — this removes the constants and leaves a pure ratio of linear terms in the new variables.

    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. Find the point where $x+y+1=0$ and $x-y-1=0$ intersect: adding gives $2x=0\Rightarrow x=0$; then $y=-1$.
    3. Shift: let $X=x$, $Y=y+1$ (so $x=X$, $y=Y-1$). Then $x+y+1=X+Y$ and $x-y-1=X-Y$.
    4. The equation becomes $\dfrac{dY}{dX}=\dfrac{X+Y}{X-Y}$ — homogeneous of degree 0.
    5. Let $Y=vX$, so $\dfrac{dY}{dX}=v+X\dfrac{dv}{dX}=\dfrac{1+v}{1-v}$.
    6. So $X\dfrac{dv}{dX}=\dfrac{1+v}{1-v}-v=\dfrac{1+v^2}{1-v}$, giving $\dfrac{1-v}{1+v^2}\,dv=\dfrac{dX}{X}$.
    7. Integrate: $\displaystyle\int\dfrac{dv}{1+v^2}-\int\dfrac{v\,dv}{1+v^2}=\log X+c$, i.e. $\tan^{-1}v-\dfrac12\log(1+v^2)=\log X+c$.
    8. Substitute back $v=Y/X=(y+1)/x$: $1+v^2=\dfrac{x^2+(y+1)^2}{x^2}$, so $\dfrac12\log(1+v^2)=\dfrac12\log(x^2+(y+1)^2)-\log|x|$.
    9. The equation becomes $\tan^{-1}\!\left(\dfrac{y+1}{x}\right)-\dfrac12\log(x^2+(y+1)^2)+\log|x|=\log|x|+c$, and the $\log|x|$ terms cancel, leaving $\tan^{-1}\!\left(\dfrac{y+1}{x}\right)-\dfrac12\log\big(x^2+y^2+2y+1\big)=c$ (expanding $(y+1)^2=y^2+2y+1$).

    Common Mistakes

    • Trying to treat the equation as homogeneous without first shifting the origin — the $+1$ constants make it non-homogeneous as originally written.
    • Losing track of the $\log|x|$ terms that cancel between the two sides.

    ✓Final answer

    The correct option is (A) — $\tan^{-1}\left(\dfrac{y+1}{x}\right)-\dfrac{1}{2}\log(x^2+y^2+2y+1)=c$.

    ANSWER: A

11. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
    Q.The substitution $x=vy$ converts which one of the following differential equation to an equation solvable by variable separable method? (A) $(y^2-2x^2y)\,dx = (x^2-2xy^2)\,dy$ (B) $x^2\,dy - y\,dx = \sqrt{x^2+y^2}\,dx$ (C) $\dfrac{dy}{dx} = \dfrac{y^2}{x+\sqrt{xy}}$ (D) $\left(1+2e^{x/y}\right) + 2e^{x/y}\left(1-\dfrac{x}{y}\right)\dfrac{dy}{dx} = 0$

    ›Reveal solutionSolution

    The equation containing $e^{x/y}$ is naturally a function of $x/y$, which is exactly what the substitution $x=vy$ (with $v=x/y$) is designed to exploit — this is the standard textbook example for that substitution.

    Concept and Intuition

    For a homogeneous differential equation, we can substitute either $y=vx$ (writing $v=y/x$, treating $x$ as the base variable) or $x=vy$ (writing $v=x/y$, treating $y$ as the base variable). The choice is guided by which ratio the equation is naturally expressed in terms of. When an equation contains a term like $e^{x/y}$, it is far more natural to substitute $x=vy$, because then $x/y=v$ directly, and $dx=v\,dy+y\,dv$ converts the whole equation into one purely in $v$ and $y$ — which then separates.

    Step-by-Step Solution

    1. Examine option (D):

    $\left(1+2e^{x/y}\right)dx + 2e^{x/y}\left(1-\frac{x}{y}\right)dy = 0$

    (written with the $dy/dx$ term brought over, equivalently $M\,dx+N\,dy=0$ form).

    2. Every term in $M$ and $N$ is a function purely of $x/y$: the equation is homogeneous, and specifically expressed via the ratio $v=x/y$.

    3. Substitute $x=vy$, so $dx = v\,dy + y\,dv$. Then:

    $(1+2e^v)(v\,dy+y\,dv) + 2e^v(1-v)\,dy = 0$

    Collect the $dy$ terms:

    $\Big[(1+2e^v)v + 2e^v(1-v)\Big]dy + (1+2e^v)y\,dv = 0$

    $\Big[v+2ve^v+2e^v-2ve^v\Big]dy + (1+2e^v)y\,dv = 0$

    $(v+2e^v)\,dy + (1+2e^v)y\,dv = 0$

    4. This is now separable in $y$ and $v$:

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

    which can be integrated directly — confirming the substitution works cleanly.

    5. Checking the other options: (A), (B), (C) are each more naturally expressed (or only genuinely homogeneous) in terms of $y/x$, so the standard $y=vx$ substitution is the natural one for them, not $x=vy$.

    Common Mistakes

    • Trying $y=vx$ on an equation written in $e^{x/y}$ form — it does not simplify, because the natural ratio here is $x/y$, not $y/x$.
    • Forgetting the product rule when substituting $x=vy$: $dx=v\,dy+y\,dv$, not just $v\,dy$.

    ✓Final answer

    The correct option is (D).

    ANSWER: D

12. 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).
    7. Substitute back $v=x+y$: $\log(x+y)=cx$ (renaming $k\to c$).

    Common Mistakes

    • Trying to solve the equation directly in $x,y$ without spotting the $dx+dy=d(x+y)$ grouping, leading to a much harder (or unsolvable-by-elementary-methods) equation.
    • Sign/inversion errors when separating $\dfrac{dv}{v\log v}=\dfrac{dx}{x}$, which could flip the roles of $x$ and $y$ and produce option (B) or (C) instead.

    ✓Final answer

    The correct option is (D) — $\log(x+y) = cx$.

    ANSWER: D