Q.Show that the differential equation $x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x$ is homogeneous and solve it.

Showing the 12 most recent of 32 on this concept.

1. 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

   $x^2\sin v\,dv=-x\,dx \ \Rightarrow\ \sin v\,dv=-\frac{dx}{x}\quad(x\neq0).$

   5. Integrate both sides: $-\cos v = -\log|x|+C_1 \Rightarrow \cos v=\log|x|-C_1=\log|x|+c$.
   6. Substituting back $v=y/x$: $\cos(y/x)=\log|x|+c$.

   Common Mistakes

   • Not spotting the cancellation of the $xv\sin v\,dx$ terms and getting stuck trying to integrate a mixed expression.
   • Sign error turning $-\cos v=-\log|x|+C_1$ into $\cos v=-\log|x|+c$ instead of $+\log|x|+c$.

   ✓Final answer

   The correct option is (A) — $\cos\left(\dfrac yx\right)=\log|x|+c$.

   ANSWER: A

2. 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: $\sin v\,dv=-\dfrac{dx}{x}$.
   5. Integrate: $-\cos v=-\log x+k$, i.e. $\log x-\cos v=-k=c$ (renaming the constant).
   6. Substitute back $v=y/x$: $\log x-\cos\dfrac{y}{x}=c$.

   Common Mistakes

   • Sign errors when moving $-\cos v$ and $-\log x$ across the equation.
   • Forgetting to substitute $v=y/x$ back at the end.

   ✓Final answer

   The correct option is (D) — $\log x-\cos\dfrac{y}{x}=c$.

   ANSWER: D

3. 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 $v=y/x$: $\cos\dfrac{y}{x}=\log_e x+c$.

   Common Mistakes

   • Forgetting that the $vx\sin v\,dx$ terms cancel — trying to solve without spotting this cancellation leads to a much harder (and wrong) equation.
   • Mixing up $y/x$ vs $x/y$ in the final answer — since the original substitution was $y=vx$, the answer must be in terms of $v=y/x$, matching option (B) rather than (A) or (C) which have $x/y$.

   ✓Final answer

   The correct option is (B) — $\cos\dfrac{y}{x}=\log_e x+c$.

   ANSWER: B

4. 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.
   4. (D) $x\dfrac{dy}{dx}=y+e^{x/y}\Rightarrow \dfrac{dy}{dx}=\dfrac{y}{x}+\dfrac{e^{x/y}}{x}$: the second term has an explicit $\dfrac1x$ outside the scale-invariant $e^{x/y}$, so under $x\to\lambda x$ this term becomes $\dfrac{e^{x/y}}{\lambda x}$, which is not equal to the original — not homogeneous.
   5. Hence only (C) qualifies.

   Common Mistakes

   • Assuming any equation containing $e^{x/y}$ is automatically homogeneous — it is only homogeneous if the rest of the expression also has matching, uniform degree.
   • Forgetting to first solve for $dy/dx$ before checking degrees (as needed in (C) and (D)).

   ✓Final answer

   The correct option is (C) — $(x^2+y^2)dx=2xy\,dy$.

   ANSWER: C

5. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\dfrac{dy}{dx} = \dfrac{y + x\tan\dfrac{y}{x}}{x} \Rightarrow \sin\dfrac{y}{x} =$ (A) $cx^2$ (B) $cx$ (C) $cx^3$ (D) $cx^4$

   ›Reveal solutionSolution

   Substituting $y=vx$ reduces the equation to $\cot v\,dv = dx/x$, giving $\sin(y/x)=cx$.

   Concept and Intuition

   An ODE of the form $dy/dx = F(y/x)$ is homogeneous and is solved by substituting $v=y/x$, which always cancels the $v$ term linearly and leaves a separable equation in $v$ and $x$.

   Step-by-Step Solution

   1. Let $y=vx \Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$.
   2. RHS: $\dfrac{y+x\tan(y/x)}{x} = v+\tan v$.
   3. So $v+x\dfrac{dv}{dx} = v+\tan v \Rightarrow x\dfrac{dv}{dx}=\tan v$.
   4. Separate: $\cot v\,dv = \dfrac{dx}{x}$.
   5. Integrate: $\log|\sin v| = \log|x|+\text{const} \Rightarrow \sin v = cx$.
   6. Substitute back $v=y/x$: $\sin(y/x)=cx$.

   Common Mistakes

   • Forgetting the $v$ terms cancel, leaving an unnecessarily complicated equation.
   • Sign/log-constant handling errors when exponentiating $\log|\sin v|=\log|x|+k$.

   ✓Final answer

   The correct option is (B) — $cx$.

   ANSWER: B

6. 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: $-\cos v = -\log|x| + C' \Rightarrow \cos v = \log|x| + C$.
   7. Substitute back $v=y/x$: $\cos(y/x) = \log x + C$.
   8. Comparing with the given general solution $\cos(y/x)=A\log x+C$: $A=1$.

   Common Mistakes

   • Sign errors when integrating $\sin v\,dv$ (the integral is $-\cos v$, easy to drop the minus sign).
   • Not simplifying/cancelling the $vx\sin v\,dx$ terms correctly, leading to an unnecessarily messy equation.

   ✓Final answer

   The correct option is (B) — 1.

   ANSWER: B

7. 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 $x/v=x^2/y$: $\log\dfrac{x^2}{y}-\dfrac{y}{x}=C$.
   9. Since $C$ is an arbitrary constant, we may equally write $-C=\log c$: negating the whole equation, $\log\dfrac{y}{x^2}+\dfrac{y}{x}=-C=\log c$, i.e. $\log\dfrac{y}{cx^2}+\dfrac{y}{x}=0$, matching the printed answer form.

   Common Mistakes

   • Sign errors when combining $v(2+v)dx$ and $v(1+v)dx$ — the whole simplification collapses to the clean $v\,dx=x(1+v)dv$ only if signs are tracked carefully.
   • Not recognising that flipping the sign of the arbitrary constant of integration is legitimate (since $c$ is arbitrary), which is needed to match the answer's printed form.

   ✓Final answer

   The correct option is (B) — $\log\left(\dfrac{y}{cx^2}\right)+\dfrac{y}{x}=0$.

   ANSWER: B

8. 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 $\mathrm{Tan}^{-1}v=\log x+\dfrac12\log(1+v^2)+C=\log\!\left(x\sqrt{1+v^2}\right)+C$.
   6. Since $v=y/x$: $x\sqrt{1+v^2}=x\sqrt{1+y^2/x^2}=\sqrt{x^2+y^2}$. So $\mathrm{Tan}^{-1}(y/x)=\log\!\left(c\sqrt{x^2+y^2}\right)$ with $c=e^C$.

   Common Mistakes

   • Losing the extra factor of $x$ when converting $\sqrt{1+v^2}$ back — it correctly cancels to leave just $\sqrt{x^2+y^2}$, no extra $x$ (ruling out the option with $cx\sqrt{x^2+y^2}$).
   • Sign error on the $\dfrac{v}{1+v^2}$ integral (it integrates to $-\tfrac12\log(1+v^2)$, i.e. it subtracts).

   ✓Final answer

   The correct option is (D) — $\mathrm{Tan}^{-1}\left(\dfrac yx\right)=\log\left(c\sqrt{x^2+y^2}\right)$.

   ANSWER: D

9. 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: $V+y\dfrac{dV}{dy} = \dfrac{g(V,1)}{f(V,1)} \Rightarrow \dfrac{dV}{dy} = \dfrac1y\left[\dfrac{g(V,1)}{f(V,1)}-V\right]$.
   6. Comparing to $\dfrac{dV}{dy}=\dfrac1y F(V)$: $F(V) = \dfrac{g(V,1)}{f(V,1)}-V$.

   Common Mistakes

   • Using $dy/dx$ directly instead of correctly switching to $dx/dy$ to match the substitution $x=Vy$ (which naturally treats $x$ as depending on $y$).
   • Mixing up whether it's $f(V,1)/g(V,1)$ or $g(V,1)/f(V,1)$ — since we inverted $dy/dx$ to get $dx/dy$, the ratio also inverts.

   ✓Final answer

   The correct option is (D) — $\left(\dfrac{g(V,1)}{f(V,1)} - V\right)$.

   ANSWER: D

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: $2\sqrt y = -2\sqrt x + C_1 \Rightarrow \sqrt x + \sqrt y = \dfrac{C_1}{2}$.
    6. Renaming the constant as $\sqrt c$ (a nonnegative constant, consistent with $\sqrt x+\sqrt y\ge0$), the general solution is $\sqrt x + \sqrt y = \sqrt c$.

    Common Mistakes

    • Trying to solve the quadratic in $p$ using the full quadratic formula without noticing the discriminant vanishes, missing the simplification.
    • Sign errors when simplifying $\sqrt{xy}/x$ to $\sqrt{y/x}$.

    ✓Final answer

    The correct option is (B) — $\sqrt{x}+\sqrt{y}=\sqrt{c}$.

    ANSWER: B

11. 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).
    7. Substitute back $v=x+y$: $\log(x+y)=cx$ (renaming $k\to c$).

    Common Mistakes

    • Trying to solve the equation directly in $x,y$ without spotting the $dx+dy=d(x+y)$ grouping, leading to a much harder (or unsolvable-by-elementary-methods) equation.
    • Sign/inversion errors when separating $\dfrac{dv}{v\log v}=\dfrac{dx}{x}$, which could flip the roles of $x$ and $y$ and produce option (B) or (C) instead.

    ✓Final answer

    The correct option is (D) — $\log(x+y) = cx$.

    ANSWER: D

12. 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 $\dfrac12\log(1+v^2)=\dfrac12\log(x^2+(y+1)^2)-\log|x|$.
    9. The equation becomes $\tan^{-1}\!\left(\dfrac{y+1}{x}\right)-\dfrac12\log(x^2+(y+1)^2)+\log|x|=\log|x|+c$, and the $\log|x|$ terms cancel, leaving $\tan^{-1}\!\left(\dfrac{y+1}{x}\right)-\dfrac12\log\big(x^2+y^2+2y+1\big)=c$ (expanding $(y+1)^2=y^2+2y+1$).

    Common Mistakes

    • Trying to treat the equation as homogeneous without first shifting the origin — the $+1$ constants make it non-homogeneous as originally written.
    • Losing track of the $\log|x|$ terms that cancel between the two sides.

    ✓Final answer

    The correct option is (A) — $\tan^{-1}\left(\dfrac{y+1}{x}\right)-\dfrac{1}{2}\log(x^2+y^2+2y+1)=c$.

    ANSWER: A