Q.Find a particular solution of the differential equation $(x - y)(dx + dy) = dx - dy$, given that $y = -1$, when $x = 0$. (Hint: put $x - y = t$)

Showing the 12 most recent of 32 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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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 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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8. 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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9. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
   Q.The general solution of the differential equation $\dfrac{dy}{dx} + \dfrac{x+y+1}{x-3y+5} = 0$ is (A) $3(y-1)^2 - 2(x+2)(y-1) - (x+2)^2 = c$ (B) $x^2 - 3y^2 - 4xy - 2x - 10y = c$ (C) $3(y+1)^2 + 2(x-2)(y+1) - (x-2)^2 = c$ (D) $x^2 + 3y^2 + 4xy + 2x + 10y = c$

   ›Reveal solutionSolution

   This is a differential equation of the form $\dfrac{dy}{dx}=-\dfrac{ax+by+c}{a'x+b'y+c'}$ with non-parallel linear terms; shifting the origin to the intersection of the two lines reduces it to a homogeneous equation. The answer is (A).

   Concept and Intuition

   When a differential equation has the form $\dfrac{dy}{dx}=-\dfrac{x+y+1}{x-3y+5}$, the numerator and denominator are linear in $x,y$ but not proportional, so it isn't directly homogeneous. The standard trick is to translate the axes to the point where the two lines $x+y+1=0$ and $x-3y+5=0$ intersect — in the new coordinates the equation becomes exactly homogeneous of degree one, which we can solve with the substitution $Y=vX$.

   Step-by-Step Solution

   1. Find the intersection point. Solve $x+y=-1$ and $x-3y=-5$ simultaneously. Subtracting: $4y=4\Rightarrow y=1$, then $x=-2$. So the lines meet at $(-2,1)$.
   2. Shift coordinates: let $X=x+2,\ Y=y-1$ (so $dX=dx,\ dY=dy$). The equation becomes $\dfrac{dY}{dX}=-\dfrac{X+Y}{X-3Y}$, which is homogeneous (every term is degree 1 in $X,Y$).
   3. Substitute $Y=vX$, so $\dfrac{dY}{dX}=v+X\dfrac{dv}{dX}$. Then

   $v+X\frac{dv}{dX}=-\frac{1+v}{1-3v}\ \Rightarrow\ X\frac{dv}{dX}=\frac{3v^2-2v-1}{1-3v}.$

   4. Separate variables:

   $\frac{1-3v}{3v^2-2v-1}\,dv=\frac{dX}{X}.$

   Factor the denominator: $3v^2-2v-1=(3v+1)(v-1)$. Partial fractions give

   $\frac{1-3v}{(3v+1)(v-1)}=\frac{-3/2}{3v+1}+\frac{-1/2}{v-1}.$

   5. Integrate both sides:

   $-\tfrac12\log|3v+1|-\tfrac12\log|v-1|=\log X + C_1$

   $\Rightarrow\ -2\log X=\log|3v+1|+\log|v-1|+C_2\ \Rightarrow\ X^{-2}=K(3v+1)(v-1).$ …

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10. 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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11. 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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12. 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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