Q.Solve the following differential equation: $x^2 dy + (xy + y^2) dx = 0; y = 1$ when $x = 1$

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 $x^3v^2=Ke^{-1/v}$. …

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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 $x/v=x^2/y$: $\log\dfrac{x^2}{y}-\dfrac{y}{x}=C$. …

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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: $\log\left(v+\sqrt{1+v^2}\right) = \log x + \log C \Rightarrow v+\sqrt{1+v^2}=Cx$. …

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4. 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 $\dfrac12\log(1+v^2)=\dfrac12\log(x^2+(y+1)^2)-\log|x|$. …

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5. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.The solution of $x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx$ when $y(\sqrt{3})=1$ is (A) $y^2+\sqrt{x^2+y^2}=x^2$ (B) $5y-\sqrt{x^2+y^2}=x^2$ (C) $y+\sqrt{x^2+y^2}=x^2$ (D) $5y^2-\sqrt{x^2+y^2}=x$

   ›Reveal solutionSolution

   A homogeneous first-order ODE; the substitution $y=vx$ reduces it to a separable equation, and the given point fixes the constant to give $y+\sqrt{x^2+y^2}=x^2$.

   Concept and Intuition

   An equation is homogeneous in $x,y$ when $\dfrac{dy}{dx}$ can be written purely as a function of $y/x$. The standard trick is $y=vx$, $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$, which always converts a homogeneous equation into a separable one in $v$ and $x$.

   Step-by-Step Solution

   1. Given $x\,dy-y\,dx=\sqrt{x^2+y^2}\,dx$, so $\dfrac{dy}{dx}=\dfrac{y+\sqrt{x^2+y^2}}{x}$ (taking $x>0$).
   2. Put $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
   3. RHS becomes $\dfrac{vx+\sqrt{x^2+v^2x^2}}{x}=v+\sqrt{1+v^2}$.
   4. So $v+x\dfrac{dv}{dx}=v+\sqrt{1+v^2}\Rightarrow x\dfrac{dv}{dx}=\sqrt{1+v^2}$.
   5. Separate: $\dfrac{dv}{\sqrt{1+v^2}}=\dfrac{dx}{x}$.
   6. Integrate: $\log\!\left(v+\sqrt{1+v^2}\right)=\log x+C_1 \Rightarrow v+\sqrt{1+v^2}=kx$ (with $k=e^{C_1}$). …

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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: $2\sqrt y = -2\sqrt x + C_1 \Rightarrow \sqrt x + \sqrt y = \dfrac{C_1}{2}$. …

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7. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.If $y=y(x)$ is the solution of $x\dfrac{dy}{dx}=y+xe^{-(y/x)}$, $y(1)=\log e$, then $y(e)=$ (A) $\log\left(\dfrac{1}{e}+1\right)$ (B) $e\log(1+e)$ (C) $e\log\left(\dfrac{1}{e}+1\right)$ (D) $e\log\left(1-\dfrac{1}{e}\right)$

   ›Reveal solutionSolution

   The ODE is homogeneous; substituting $y=vx$ separates variables to $e^v\,dv=dx/x$, and applying the initial condition $y(1)=1$ gives $y(e)=e\log(1+e)$.

   Concept and Intuition

   $x\dfrac{dy}{dx}=y+xe^{-y/x}$ is homogeneous of degree $0$ in $x,y$ (both terms on the right, divided by $x$, depend only on $y/x$). The standard substitution $y=vx$ always reduces such equations to a separable one in $v$ and $x$.

   Step-by-Step Solution

   1. Let $y=vx$, so $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
   2. Substitute: $x\left(v+x\dfrac{dv}{dx}\right)=vx+xe^{-v}\Rightarrow x^2\dfrac{dv}{dx}=xe^{-v}\Rightarrow x\dfrac{dv}{dx}=e^{-v}$.
   3. Separate: $e^v\,dv=\dfrac{dx}{x}$. Integrate: $e^v=\log x+C$.
   4. Initial condition: $y(1)=\log e=1$, so $v=y/x=1$ at $x=1$: $e^1=\ln1+C\Rightarrow e=C$.
   5. So $e^{y/x}=\log x+e$. …

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8. 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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9. 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 $V=y/x$: $-\dfrac{x}{y}-\log\dfrac{y}{x}=\log x+C_1 \Rightarrow -\dfrac{x}{y}-\log y+\log x=\log x+C_1 \Rightarrow -\dfrac{x}{y}-\log y=C_1$. …

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10. 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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11. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.If $\cos\dfrac{y}{x} = A\log x + C$ is the general solution of $\left(x\sin\dfrac{y}{x}\right) dy = \left(y\sin\dfrac{y}{x} - x\right) dx$, then $A =$ (A) 2 (B) 1 (C) -1 (D) -2

    ›Reveal solutionSolution

    The equation is homogeneous; the substitution $y=vx$ reduces it to a separable form whose integration directly gives $\cos(y/x)=\log x + C$, so $A=1$.

    Concept and Intuition

    The differential equation $(x\sin(y/x))dy=(y\sin(y/x)-x)dx$ is homogeneous (every term scales the same way under $x\to\lambda x, y\to\lambda y$), so the standard substitution $y=vx$ (with $v=y/x$) converts it into a separable equation in $v$ and $x$.

    Step-by-Step Solution

    1. Let $y=vx$, so $dy=v\,dx+x\,dv$.
    2. Substitute into $(x\sin v)dy = (vx\sin v - x)dx$: $(x\sin v)(v\,dx+x\,dv) = (vx\sin v-x)dx$.
    3. Expand LHS: $vx\sin v\,dx + x^2\sin v\,dv = vx\sin v\,dx - x\,dx$.
    4. The $vx\sin v\,dx$ terms cancel from both sides: $x^2\sin v\,dv = -x\,dx$.
    5. Divide by $x^2$ (assuming $x\ne0$): $\sin v\,dv = -\dfrac{dx}{x}$.
    6. Integrate both sides: $-\cos v = -\log|x| + C' \Rightarrow \cos v = \log|x| + C$. …

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12. 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 $\dfrac1v=\dfrac{x}{y}$: $\log(xy)+\dfrac{x}{y}=K$. …

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