Q.Show that the differential equation $2y\,e^{x/y}\,dx + \left(y - 2x\,e^{x/y}\right)dy = 0$ is homogeneous and find its particular solution, given that, $x = 0$ when $y = 1$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Which one of the following is a homogeneous differential equation? (A) $\dfrac{dy}{dx}=x^3+(\sin x)y$ (B) $\dfrac{dy}{dx}=(x^3+y^3)e^{x/y}+x\sqrt{y}$ (C) $(x^2+y^2)dx=2xy\,dy$ (D) $x\dfrac{dy}{dx}=y+e^{x/y}$

   ›Reveal solutionSolution

   Only option (C) has both numerator and denominator of the same total degree when written as $dy/dx=F(x,y)$, so it is the homogeneous equation.

   Concept and Intuition

   A first-order DE $\dfrac{dy}{dx}=F(x,y)$ is homogeneous if $F(\lambda x,\lambda y)=F(x,y)$ for all $\lambda$ — equivalently, if $F$ can be written purely as a function of $t=y/x$. The quickest test is to check whether every additive term in $F$ scales with the same power of $\lambda$ when $x\to\lambda x, y\to\lambda y$.

   Step-by-Step Solution

   1. (A) $\dfrac{dy}{dx}=x^3+(\sin x)y$: under $x\to\lambda x,y\to\lambda y$, this becomes $\lambda^3x^3+(\sin\lambda x)\lambda y$ — the terms don't scale uniformly, and $\sin x$ isn't scale-invariant at all. Not homogeneous.
   2. (B) $\dfrac{dy}{dx}=(x^3+y^3)e^{x/y}+x\sqrt y$: the exponential factor $e^{x/y}$ is scale-invariant (since $x/y$ is unchanged by $x\to\lambda x,y\to\lambda y$), so $(x^3+y^3)e^{x/y}$ scales as $\lambda^3$; but $x\sqrt y$ scales as $\lambda^{3/2}$. Different degrees $\Rightarrow$ not homogeneous.
   3. (C) Rewrite as $\dfrac{dy}{dx}=\dfrac{x^2+y^2}{2xy}$. Numerator scales as $\lambda^2$, denominator as $\lambda^2$, so the ratio is invariant under $x\to\lambda x,y\to\lambda y$ — genuinely a function of $y/x$ only: $\dfrac{1+(y/x)^2}{2(y/x)}$. Homogeneous. …

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2. 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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3. 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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4. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.Suppose that f(x,y) and g(x,y) are homogeneous functions of same order. If x=Vy reduces the equation $\dfrac{dy}{dx} = \dfrac{f(x,y)}{g(x,y)}$ to the form $\dfrac{dV}{dy} = \dfrac{1}{y}(F(V))$, then $F(V) =$ (A) $\left(\dfrac{f(1,V)}{g(1,V)} - V\right)$ (B) $\left(\dfrac{f(V,1)}{g(V,1)} - V\right)$ (C) $\left(\dfrac{g(1,V)}{f(1,V)} - V\right)$ (D) $\left(\dfrac{g(V,1)}{f(V,1)} - V\right)$

   ›Reveal solutionSolution

   Using $x=Vy$ and inverting the given ODE to work with $dx/dy$ (not $dy/dx$) shows $F(V)=\dfrac{g(V,1)}{f(V,1)}-V$.

   Concept and Intuition

   This is the homogeneous-equation substitution, but with the roles of $x$ and $y$ swapped compared to the usual $y=Vx$ textbook form (here it's $x=Vy$, and the new variable is a function of $y$, not $x$). Since a homogeneous function of degree $n$ satisfies $h(x,y)=y^n h(x/y,1)=y^n h(V,1)$, ratios of two same-order homogeneous functions simplify by cancelling the $y^n$ factor.

   Step-by-Step Solution

   1. Given $\dfrac{dy}{dx}=\dfrac{f(x,y)}{g(x,y)}$, so its reciprocal is $\dfrac{dx}{dy}=\dfrac{g(x,y)}{f(x,y)}$.
   2. Let $x=Vy$ (V a function of y). Then $\dfrac{dx}{dy} = V + y\dfrac{dV}{dy}$ (product rule).
   3. Since $f,g$ are homogeneous of the same degree $n$: $f(x,y)=f(Vy,y)=y^n f(V,1)$ and $g(x,y)=y^n g(V,1)$.
   4. So $\dfrac{g(x,y)}{f(x,y)} = \dfrac{y^n g(V,1)}{y^n f(V,1)} = \dfrac{g(V,1)}{f(V,1)}$.
   5. Equate: $V+y\dfrac{dV}{dy} = \dfrac{g(V,1)}{f(V,1)} \Rightarrow \dfrac{dV}{dy} = \dfrac1y\left[\dfrac{g(V,1)}{f(V,1)}-V\right]$. …

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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 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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7. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.If $\dfrac{dy}{dx}=f(x,y)$ is a homogeneous differential equation, then the general form of $f(x,y)$ is (A) $x^n \phi\left(\dfrac{y}{x}\right), n\ne 1$ (B) $y^n \phi\left(\dfrac{x}{y}\right), n\ne 1$ (C) $\phi\left(\dfrac{y}{x}\right)$ (D) $K^n f(x,y), n\ne 1$

   ›Reveal solutionSolution

   By definition, the DE $dy/dx=f(x,y)$ is called homogeneous exactly when $f$ is homogeneous of degree zero, which forces the general form $f(x,y)=\phi(y/x)$ (no leading power of $x$).

   Concept and Intuition

   A function $F(x,y)$ is homogeneous of degree $n$ if $F(\lambda x,\lambda y)=\lambda^n F(x,y)$ for every nonzero $\lambda$; such a function can always be written in the form $x^n\phi(y/x)$ (factor out $x^n$ and what remains depends only on the ratio $y/x$). The differential equation $dy/dx=f(x,y)$ is specifically called "homogeneous" when $f$ itself is homogeneous of degree zero — this is what guarantees that substituting $y=vx$ turns the equation into one separable in $v$ and $x$.

   Step-by-Step Solution

   1. Recall the general form of a degree-$n$ homogeneous function:

   $F(x,y) = x^n\,\phi\!\left(\frac{y}{x}\right)$

   2. For the differential equation $\dfrac{dy}{dx}=f(x,y)$ to qualify as "homogeneous" in the standard sense used to justify the $y=vx$ substitution, we require $f(\lambda x,\lambda y) = f(x,y)$ for all $\lambda\ne0$ — i.e. degree $n=0$.

   3. Substituting $n=0$ into the general form:

      $f(x,y) = x^0\,\phi\!\left(\frac{y}{x}\right) = \phi\!\left(\frac{y}{x}\right)$ …

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8. 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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9. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.If $x^\alpha \dfrac{dy}{dx} = y^\beta(\gamma\log x + \delta\log y + 1)$ is a homogeneous differential equation, then (A) $\alpha=\beta$ and $\gamma=-\delta$ (B) $\alpha=\beta$ and $\gamma=\delta$ (C) $\alpha\ne\beta$ and $\gamma=\delta$ (D) $\alpha\ne\beta$ and $\gamma\ne\delta$

   ›Reveal solutionSolution

   Testing the differential equation under the scaling $x\to\lambda x, y\to\lambda y$ and demanding invariance (the defining property of a homogeneous ODE) forces $\alpha=\beta$ and $\gamma=-\delta$.

   Concept and Intuition

   A first-order ODE $dy/dx=F(x,y)$ is homogeneous exactly when $F(\lambda x,\lambda y)=F(x,y)$ for every $\lambda>0$ — i.e. $dy/dx$ depends only on the ratio $y/x$. Substituting the scaling directly into the given equation and demanding this invariance for all $\lambda$ pins down the required conditions on the exponents/coefficients.

   Step-by-Step Solution

   1. Rewrite: $\dfrac{dy}{dx} = \dfrac{y^\beta(\gamma\log x+\delta\log y+1)}{x^\alpha}$.
   2. Substitute $x\to\lambda x,\ y\to\lambda y$: $\dfrac{(\lambda y)^\beta\big[\gamma\log(\lambda x)+\delta\log(\lambda y)+1\big]}{(\lambda x)^\alpha}$ $=\lambda^{\beta-\alpha}\cdot\dfrac{y^\beta\big[(\gamma+\delta)\log\lambda + \gamma\log x+\delta\log y+1\big]}{x^\alpha}$.
   3. For this to equal the original expression for every $\lambda$, two things must vanish: (i) the overall power $\lambda^{\beta-\alpha}$ must be $1$ for all $\lambda$ $\Rightarrow \alpha=\beta$; …

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10. 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 $v$: …

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11. 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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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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