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 32 on this concept.

1. 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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2. 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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3. 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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4. 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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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 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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7. 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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8. 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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9. 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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10. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.If $X=x+h$, $Y=y+k$ transforms $\dfrac{dy}{dx}=\dfrac{2x+3y-7}{3x+2y-8}$ to a homogeneous differential equation, then $(h,k)=$ (A) $(1,2)$ (B) $(2,1)$ (C) $(7,8)$ (D) $(8,7)$

    ›Reveal solutionSolution

    To make $\frac{dy}{dx}=\frac{2x+3y-7}{3x+2y-8}$ homogeneous, shift the origin to the point where both linear expressions vanish simultaneously; solving the two equations gives $(h,k)=(2,1)$.

    Concept and Intuition

    A differential equation of the form $\dfrac{dy}{dx}=\dfrac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2}$ (with nonzero constants $c_1,c_2$) is made homogeneous by translating the origin to the point $(h,k)$ that is the simultaneous solution of $a_1x+b_1y+c_1=0$ and $a_2x+b_2y+c_2=0$ — this removes the constant terms and leaves purely ratio-of-linear-forms in the new variables $X=x-h,\,Y=y-k$ (or $x=X+h,\,y=Y+k$ as stated), which is homogeneous of degree $0$.

    Step-by-Step Solution

    1. We need $(h,k)$ such that both numerator and denominator vanish at $x=h,\,y=k$: $2h+3k-7=0$ and $3h+2k-8=0$.
    2. Rewrite: $2h+3k=7$ …(i), $3h+2k=8$ …(ii).
    3. Multiply (i) by $2$: $4h+6k=14$. Multiply (ii) by $3$: $9h+6k=24$. …

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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-AN1 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) $\cos\dfrac{x}{y} = \log_e x + c$ (B) $\cos\dfrac{y}{x} = \log_e x + c$ (C) $\cos\dfrac{x}{y} = \log_e y + c$ (D) $\cos\dfrac{y}{x} = \log_e y + c$

    ›Reveal solutionSolution

    This is a homogeneous DE; the substitution $y=vx$ cleanly separates variables, giving $\cos(y/x)=\log_e x+c$ — (B).

    Concept and Intuition

    An equation is homogeneous when every term has the same total "degree" once you notice it depends only on the ratio $y/x$. The standard substitution $y=vx$ turns it into a separable equation in $v$ and $x$.

    Step-by-Step Solution

    1. Given: $\left(x\sin\dfrac{y}{x}\right)dy = \left(y\sin\dfrac{y}{x}-x\right)dx$.
    2. Let $y=vx\Rightarrow dy=v\,dx+x\,dv$, and $\dfrac{y}{x}=v$.
    3. Substitute: $x\sin v\,(v\,dx+x\,dv) = (vx\sin v - x)\,dx$.
    4. Expand LHS: $vx\sin v\,dx+x^2\sin v\,dv$. RHS: $vx\sin v\,dx - x\,dx$.
    5. The $vx\sin v\,dx$ term cancels from both sides, leaving: $x^2\sin v\,dv=-x\,dx$.
    6. Divide by $x^2$ (assuming $x\ne0$): $\sin v\,dv=-\dfrac{dx}{x}$.
    7. Integrate both sides: $-\cos v=-\ln|x|+C_1\Rightarrow\cos v=\ln|x|+C$ (relabeling the constant).
    8. Substitute back $v=y/x$: $\cos\dfrac{y}{x}=\log_e x+c$. …

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