Q.A homogeneous differential equation of the form $\dfrac{dx}{dy} = h\left(\dfrac{x}{y}\right)$ can be solved by making the substitution (A) $y = vx$ (B) $v = yx$ (C) $x = vy$ (D) $x = v$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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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-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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8. 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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9. AP EAPCET 2025Set eng-2025-05-27-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) $\cos\left(\dfrac{y}{x}\right) = \log|x| + c$ (B) $\cos\left(\dfrac{y}{x}\right) = \dfrac{1}{x} + c$ (C) $\cos\left(\dfrac{x}{y}\right) = \log|y| + c$ (D) $\cos\dfrac{y}{x} = \dfrac{2}{x} + c$

   ›Reveal solutionSolution

   The equation is homogeneous in $x,y$; the substitution $y=vx$ makes it separable and integrates to $\cos(y/x)=\log|x|+c$.

   Concept and Intuition

   A differential equation where every term has the same total degree in $x,y$ (here, degree 1, since $x\sin(y/x)$ and $y\sin(y/x)$ and $x$ are all "degree 1" once $y/x$ is treated as dimensionless) is homogeneous, and the standard move is $y=vx$: it reduces the two variables $x,y$ to one variable $v=y/x$ plus $x$, making the equation separable.

   Step-by-Step Solution

   1. Given: $(x\sin(y/x))\,dy=(y\sin(y/x)-x)\,dx$.
   2. Let $y=vx$, so $dy=v\,dx+x\,dv$ and $y/x=v$. Substitute:

   $x\sin v\,(v\,dx+x\,dv) = (vx\sin v-x)\,dx.$

   3. Expand: $xv\sin v\,dx+x^2\sin v\,dv = xv\sin v\,dx-x\,dx$.
   4. The $xv\sin v\,dx$ terms on both sides cancel exactly, leaving $x^2\sin v\,dv=-x\,dx \ \Rightarrow\ \sin v\,dv=-\frac{dx}{x}\quad(x\neq0).$ …

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10. 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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11. 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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12. 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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