Q.Solve the following differential equation: $x dy - y dx = \sqrt{x^2 + y^2} dx$

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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 $\mathrm{Tan}^{-1}v=\log x+\dfrac12\log(1+v^2)+C=\log\!\left(x\sqrt{1+v^2}\right)+C$. …

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6. 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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7. 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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8. 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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9. 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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10. 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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11. 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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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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