Q.Find the particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = 3x + 4y$ given that $y = 0$ when $x = 0$.

Showing the 12 most recent of 13 on this concept.

1. COMEDK 2025Set 2025-E1 markMCQ
   Q.The solution of the differential equation: $x \cos y d y=\left(x e^x \log x+e^x\right) d x$ is (A) $\sin y-e^x \log x=c$ (B) $\sin y=e^x \log x+c$ (C) $\sin y=e^x+\log x+c$ (D) $\sin y=\frac{e^x}{x}+c$

   ›Reveal solutionSolution

   The differential equation is separable after rewriting, leading to $\sin y = e^x \log x + c$, which matches option (B).

   We start with the given equation:

   $x \cos y \, dy = \left( x e^x \log x + e^x \right) dx$

   The key idea is to separate variables — get all $y$ terms with $dy$ and all $x$ terms with $dx$. This works because the equation is already in a form where the left side depends only on $y$ and the right side only on $x$, once we divide appropriately.

   1. Divide both sides by $x$ (assuming $x \neq 0$):

   $\cos y \, dy = \frac{x e^x \log x + e^x}{x} \, dx$

   2. Simplify the right-hand side:

   $\frac{x e^x \log x + e^x}{x} = e^x \log x + \frac{e^x}{x}$

   So the equation becomes:

   $\cos y \, dy = \left( e^x \log x + \frac{e^x}{x} \right) dx$

   3. Integrate both sides:

      • Left side: $\int \cos y \, dy = \sin y + C_1$
      • Right side: $\int \left( e^x \log x + \frac{e^x}{x} \right) dx$

      Notice that the right side is the derivative of $e^x \log x$ with respect to $x$:

   $\frac{d}{dx} \left( e^x \log x \right) = e^x \log x + e^x \cdot \frac{1}{x}$

   This is exactly our integrand.

   4. Therefore:

   $\int \left( e^x \log x + \frac{e^x}{x} \right) dx = e^x \log x + C_2$

   5. Combine constants:

   $\sin y = e^x \log x + c$

   where $c = C_2 - C_1$ is an arbitrary constant.

   Tip

   Spotting that $e^x \log x + \frac{e^x}{x}$ is the derivative of $e^x \log x$ saves time — it’s a classic product rule pattern.

   Watch out

   A common mistake is to try integrating $e^x \log x$ and $\frac{e^x}{x}$ separately, not noticing they come from the same derivative. That leads to unnecessary complexity.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. COMEDK 2025Set 2025-M1 markMCQ
   Q.The solution of the differential equation $\frac{d y}{d x}+y \log y \cot x=0$ is (A) $\cos x \log y=c$ (B) $\log y=c \sin x$ (C) $\sin x \log y=c$ (D) $y \sin x=c$

   ›Reveal solutionSolution

   This is a first-order separable ODE. By separating variables and integrating, we find that $\sin x \log y = c$, so the correct option is (C).

   We start with the differential equation

   $\frac{dy}{dx} + y \log y \cot x = 0.$

   The key insight is that this equation is separable: we can rearrange it so that all terms involving $y$ (and $dy$) are on one side, and all terms involving $x$ (and $dx$) are on the other. The presence of $y \log y$ suggests a substitution or direct separation.

   1. Rewrite the equation Bring the second term to the right-hand side:

   $\frac{dy}{dx} = - y \log y \cot x.$

   Now separate variables by dividing both sides by $y \log y$ and multiplying by $dx$:

   $\frac{dy}{y \log y} = - \cot x \, dx.$

   2. Integrate both sides The left side is a standard logarithmic integral. Let $u = \log y$, then $du = \frac{dy}{y}$, so

   $\int \frac{dy}{y \log y} = \int \frac{du}{u} = \log |u| = \log |\log y|.$

   The right side: recall $\cot x = \frac{\cos x}{\sin x}$, so

   $\int -\cot x \, dx = - \int \frac{\cos x}{\sin x} dx = - \log |\sin x| + C.$

   (Here we used the substitution $v = \sin x$, $dv = \cos x \, dx$.)

   3. Combine constants Equating the integrals:

   $\log |\log y| = - \log |\sin x| + C.$

   Exponentiate both sides:

   $|\log y| = e^{- \log |\sin x| + C} = e^C \cdot e^{\log |\sin x|^{-1}} = \frac{e^C}{|\sin x|}.$

   Let $c_1 = \pm e^C$ (absorbing the absolute value), we get

   $\log y = \frac{c_1}{\sin x}.$

   Multiply both sides by $\sin x$:

   $\sin x \log y = c_1.$

   Since $c_1$ is an arbitrary constant, we rename it $c$.

   4. Check the form The result is $\sin x \log y = c$, which matches option (C).

   Watch out

   A common mistake is to forget the absolute values or to mis-integrate $\cot x$ as $\log |\sin x|$ without the negative sign. Also, note that $\log y$ here means natural logarithm, but the constant absorbs any base change.

   Tip

   If you see $y \log y$ in a differential equation, immediately think of the substitution $u = \log y$ — it turns the left side into $\frac{du}{u}$, a clean logarithmic integral.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

3. COMEDK 2022Set 20221 markMCQ
   Q.The solution of the differential equation $x \frac{d y}{d x}=\cot y$ is (A) $y \cos x=c$ (B) $x \cos y=c$ (C) $\log (x \cos y)=c$ (D) $\log (y \cos x)=c$

   ›Reveal solutionSolution

   The differential equation is separable: rewrite as $x\,dy = \cot y\,dx$, separate variables, integrate, and simplify to $x\cos y = c$, which matches option (B).

   We start with the given differential equation:

   $x \frac{dy}{dx} = \cot y$

   The key insight is that this is a separable first-order ODE — we can rearrange it so that all $y$ terms (including $dy$) are on one side and all $x$ terms (including $dx$) are on the other. Once separated, we integrate both sides and then solve for the relationship between $x$ and $y$.

   1. Separate the variables Multiply both sides by $dx$ and divide by $x$ and $\cot y$ (but it's easier to write $\cot y = \frac{\cos y}{\sin y}$ and treat it directly):

   $x\,dy = \cot y \, dx \quad\Rightarrow\quad \frac{dy}{\cot y} = \frac{dx}{x}$

   Since $\frac{1}{\cot y} = \tan y$, this becomes:

   $\tan y \, dy = \frac{dx}{x}$

   2. Integrate both sides

   $\int \tan y \, dy = \int \frac{dx}{x}$

   The integral of $\tan y$ is $-\log|\cos y|$ (because $\frac{d}{dy}(-\log|\cos y|) = \tan y$), and the integral of $\frac{1}{x}$ is $\log|x|$. So:

   $-\log|\cos y| = \log|x| + C$

   3. Combine the logarithms Multiply both sides by $-1$:

   $\log|\cos y| = -\log|x| - C$

   Write $-\log|x| = \log|x^{-1}| = \log\left|\frac{1}{x}\right|$, and let $C' = -C$ (a new constant):

   $\log|\cos y| = \log\left|\frac{1}{x}\right| + C'$

   Exponentiate both sides:

   $|\cos y| = e^{C'} \cdot \frac{1}{|x|}$

   Since $e^{C'}$ is a positive constant, we can rename it as $k > 0$, giving:

   $|\cos y| = \frac{k}{|x|}$

   Removing absolute values (the constant can absorb the sign), we get:

   $\cos y = \frac{c}{x}$

   where $c$ is an arbitrary constant (positive or negative).

   4. Rewrite in the form given in the options Multiply both sides by $x$:

   $x \cos y = c$

   This matches option (B) exactly.

   Watch out

   A common mistake is to incorrectly integrate $\tan y$ as $\log|\sin y|$ or to forget the minus sign. Always remember: $\int \tan y\,dy = -\log|\cos y| + C$.

   Tip

   You can check the answer quickly by differentiating $x\cos y = c$ implicitly:

   $\cos y\,dx - x\sin y\,dy = 0 \Rightarrow x\sin y\,dy = \cos y\,dx \Rightarrow x\frac{dy}{dx} = \frac{\cos y}{\sin y} = \cot y$, which is exactly the original equation.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

4. COMEDK 2024Set 2024-M1 markMCQ
   Q.The particular solution of $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$, when $x=0, y=\frac{1}{2}$ is (A) $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{6}$ (B) $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{2}$ (C) $\sin ^{-1} x-\sin ^{-1} y=\frac{\pi}{2}$ (D) $\sin ^{-1} x+\sin ^{-1} y=\frac{\pi}{3}$

   ›Reveal solutionSolution

   The given differential equation is separable and leads to the relation $\sin^{-1} x + \sin^{-1} y = C$. Using the initial condition $x=0, y=1/2$ gives $C = \pi/6$, so the particular solution is $\sin^{-1} x + \sin^{-1} y = \pi/6$, which corresponds to option (A).

   Concept & Intuition

   The equation is of the form $y' + \sqrt{\frac{1-y^2}{1-x^2}} = 0$. The square root suggests a connection to inverse trigonometric derivatives: recall that $\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}$, and similarly for $y$. This hints that separating variables will let us integrate directly to a sum of arcsines. The initial condition then pins down the constant.

   Step-by-step solution

   1. Rewrite the equation

   $\frac{dy}{dx} = -\sqrt{\frac{1-y^2}{1-x^2}}$

   The negative sign tells us that as $x$ increases, $y$ decreases (or vice versa), consistent with an inverse sine sum being constant.

   2. Separate variables Multiply both sides by $dx$ and divide by $\sqrt{1-y^2}$ (valid where $|y|<1$, which holds near $y=1/2$):

   $\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$

   3. Integrate both sides The left side integrates to $\sin^{-1} y$, the right side to $-\sin^{-1} x$:

   $\int \frac{dy}{\sqrt{1-y^2}} = -\int \frac{dx}{\sqrt{1-x^2}}$

   $\sin^{-1} y = -\sin^{-1} x + C$

   Rearranging:

   $\sin^{-1} x + \sin^{-1} y = C$

   4. Apply the initial condition At $x=0$, $y=\frac12$:

   $\sin^{-1}(0) + \sin^{-1}\!\left(\frac12\right) = C$

   Since $\sin^{-1}(0)=0$ and $\sin^{-1}(1/2)=\pi/6$, we get:

   $C = \frac{\pi}{6}$

   5. Write the particular solution

   $\sin^{-1} x + \sin^{-1} y = \frac{\pi}{6}$

   This matches option (A).

   Tip

   A common mistake is forgetting the negative sign when separating, which would give $\sin^{-1} x - \sin^{-1} y = C$ — leading to a wrong constant. Always check the sign by plugging the initial condition into the candidate.

   Watch out

   The domain of $\sin^{-1}$ is $[-1,1]$, and the initial point $(0, 1/2)$ is safely inside. The solution curve stays within this region as long as $|x| \le 1$ and $|y| \le 1$, which is guaranteed by the arcsine relation.

   ✓Final answer

   The correct option is (A).

   ANSWER: A

5. COMEDK 2024Set 2024-E1 markMCQ
   Q.

   [!FORMULA] $\text { If } \frac{d y}{d x}=y+3>0 \text { and } y(0)=2 \text { then } y(\log 2) \text { is equal to }$

   (A) 5 (B) 13 (C) $-$2 (D) 7

   ›Reveal solutionSolution

   This is a first-order linear ODE solved by separation of variables; using the initial condition gives $y = 5e^x - 3$, so $y(\log 2) = 7$, which corresponds to option (D).

   We are given $\frac{dy}{dx} = y + 3$ with $y+3 > 0$ and $y(0) = 2$. The condition $y+3 > 0$ ensures we never divide by zero or take logs of a negative number — a helpful safety net.

   Concept & Intuition

   The equation says: the rate of change of $y$ is proportional to how far $y$ is above $-3$. This is classic exponential growth (or decay) toward or away from a constant. Here, since $y(0)=2 > -3$, the quantity $y+3$ will grow exponentially. We can solve by separating variables — put all $y$ terms on one side, $x$ terms on the other, then integrate.

   Step-by-step solution

   1. Separate variables Write $\frac{dy}{dx} = y+3$ as

   $\frac{dy}{y+3} = dx.$

   This is valid because $y+3 > 0$ ensures no division by zero.

   2. Integrate both sides

   $\int \frac{dy}{y+3} = \int dx \quad\Rightarrow\quad \log|y+3| = x + C.$

   Since $y+3 > 0$, we can drop the absolute value:

   $\log(y+3) = x + C.$

   3. Solve for $y$ Exponentiate:

   $y+3 = e^{x+C} = e^C e^x.$

   Let $A = e^C$, so $y = A e^x - 3$.

   4. Use the initial condition $y(0)=2$ Substitute $x=0$, $y=2$:

   $2 = A e^0 - 3 \quad\Rightarrow\quad 2 = A - 3 \quad\Rightarrow\quad A = 5.$

   Hence the particular solution is

   $y(x) = 5e^x - 3.$

   5. Evaluate at $x = \log 2$

   $y(\log 2) = 5 e^{\log 2} - 3 = 5 \cdot 2 - 3 = 10 - 3 = 7.$

   Tip

   Notice that $e^{\log 2} = 2$ exactly — no calculator needed. This is a common trick in such problems.

   Watch out

   A classic mistake is forgetting the constant of integration or misapplying the initial condition. Always check that your solution satisfies $y(0)=2$ before plugging in the final $x$.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

6. KCET 2025Set A-11 markMCQ
   Q.General solution of the differential equation $\frac{dy}{dx} + y \tan x = \sec x$ is (A) $y \sec x = \tan x + c$ (B) $y \tan x = \sec x + c$ (C) $\cos x \sec x = y \tan x + c$ (D) $x \sec x = \tan y + c$

   ›Reveal solutionSolution

   Recognise a first-order linear ODE, build the integrating factor $\sec x$, and integrate $\sec^2 x$.

   Step 1 — Identify the form.

   The equation

   $\frac{dy}{dx} + y\tan x = \sec x$

   is of the standard linear type $\dfrac{dy}{dx} + P(x)\,y = Q(x)$ with

   $P(x) = \tan x, \qquad Q(x) = \sec x .$

   Step 2 — Why an integrating factor works.

   Multiplying a linear ODE by $\mu(x)=e^{\int P\,dx}$ turns the whole left side into an exact derivative $\frac{d}{dx}\big(\mu y\big)$, because $\mu' = \mu P$. That is the entire trick — it converts the ODE into a straight integration.

   Step 3 — Compute the integrating factor.

   $\mu = e^{\int \tan x\,dx} = e^{\log|\sec x|} = \sec x$

   Step 4 — Multiply and collapse the left side.

   $\sec x\,\frac{dy}{dx} + y\sec x\tan x = \sec^2 x$

   The left side is exactly $\dfrac{d}{dx}\big(y\sec x\big)$, since $\dfrac{d}{dx}(\sec x)=\sec x\tan x$.

   $\frac{d}{dx}\big(y\sec x\big) = \sec^2 x$

   Step 5 — Integrate both sides.

   $y\sec x = \int \sec^2 x\,dx = \tan x + c$

   Step 6 — Check. Differentiate $y\sec x=\tan x+c$: $y'\sec x + y\sec x\tan x=\sec^2 x$. Dividing by $\sec x$ returns $y' + y\tan x=\sec x$ ✓.

   ✓Final answer

   The correct option is (A) $y \sec x = \tan x + c$.

   ANSWER: A

7. COMEDK 2021Set 20211 markMCQ
   Q.The solution of the differential equation ${\sec ^2}x\tan ydx + {\sec ^2}y\tan xdy = 0$ is (A) $\tan y\,.\,\tan x = C$ (B) ${{\tan y} \over {\tan x}} = C$ (C) ${{{{\tan }^2}x} \over {\tan y}} = C$ (D) None of these

   ›Reveal solutionSolution

   log|tan x| + log|tan y| = log C => log|tan x * tan y| = log C => tan x * tan y = C.

   Concept: variables separable differential equation.

   sec^2(x) tan(y) dx + sec^2(y) tan(x) dy = 0

   Divide throughout by tan(x) tan(y):

   [sec^2(x)/tan(x)] dx + [sec^2(y)/tan(y)] dy = 0.

   Integrate. Note that d(tan x) = sec^2(x) dx, so integral of sec^2(x)/tan(x) dx = log|tan x|.

   log|tan x| + log|tan y| = log C

   => log|tan x * tan y| = log C

   => tan x * tan y = C.

   ✓Final answer

   The correct option is (A) — $\tan y\,.\,\tan x = C$

   ANSWER: A

8. COMEDK 2022Set 20221 markMCQ
   Q.The solution of the differential equation $\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}=0$ is (A) $\cos ^{-1} x+\cos ^{-1} y=c$ (B) $\sin ^{-1} x+\sin ^{-1} y=c$ (C) $\cosh ^{-1} x+\cosh ^{-1} y=c$ (D) $\sinh ^{-1} x+\sinh ^{-1} y=c$

   ›Reveal solutionSolution

   Separating the variables gives $\frac{dy}{\sqrt{1-y^2}}=-\frac{dx}{\sqrt{1-x^2}}$; integrating yields $\sin^{-1}x+\sin^{-1}y=c$. The correct option is (B).

   Concept

   The equation is variables-separable: all $y$-terms move to one side and all $x$-terms to the other. The standard integral $\int\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t$ (not an inverse-hyperbolic function, which would need $\sqrt{1+t^2}$ or $\sqrt{t^2-1}$) then gives the answer directly.

   Solution

   1. Rearrange. $\dfrac{dy}{dx}=-\dfrac{\sqrt{1-y^2}}{\sqrt{1-x^2}}$.
   2. Separate. $\dfrac{dy}{\sqrt{1-y^2}}=-\dfrac{dx}{\sqrt{1-x^2}}$.
   3. Integrate. $\sin^{-1}y=-\sin^{-1}x+c$.
   4. Rearrange. $\sin^{-1}x+\sin^{-1}y=c$.

   Tip

   The inverse-hyperbolic forms in (C) and (D) arise from $\sqrt{1+x^2}$ or $\sqrt{x^2-1}$, not $\sqrt{1-x^2}$; the directly integrated standard form here is the inverse-sine one.

   ✓Final answer

   The correct option is (B) — $\sin^{-1}x+\sin^{-1}y=c$.

9. COMEDK 2023Set 2023-M1 markMCQ
   Q.The solution of the differential equation $\left(\frac{d y}{d x}\right) \tan y=\sin (x+y)+\sin (x-y)$ is (A) $\sec x=-2 \sec y+C$ (B) $\sec y=2 \cos y+C$ (C) $\sec y=-2 \cos x+C$ (D) $\sec x=-2 \cos y+C$

   ›Reveal solutionSolution

   $\sin(x+y)+\sin(x-y)=2\sin x\cos y$; separating gives $\sec y=-2\cos x+C$.

   Use $\sin(x+y)+\sin(x-y)=2\sin x\cos y$. The equation becomes

   $\frac{dy}{dx}\tan y=2\sin x\cos y\;\Rightarrow\;\frac{\sin y}{\cos y}\,dy=2\sin x\cos y\,dx.$

   Divide by $\cos y$:

   $\frac{\sin y}{\cos^2 y}\,dy=2\sin x\,dx.$

   Integrate the left side with $u=\cos y,\;du=-\sin y\,dy$: $\int\frac{-du}{u^2}=\frac1u=\sec y$. The right side gives $-2\cos x$:

   $\sec y=-2\cos x+C.$

   ✓Final answer

   The correct option is (C) — $\sec y=-2 \cos x+C$

10. COMEDK 2021Set 2021-B1 markMCQ
    Q.The general solution of $x + y\frac{dy}{dx} = \sec(x^2 + y^2)$ is (A) $\sin(x^2 + y^2) = 2x + c$ (B) $-\sin(x^2 + y^2) = x/2 + c$ (C) $\sin(x^2 + y^2) = 2x + c$ (D) $-\sin(x^2 + y^2) = 2x + c$

    ›Reveal solutionSolution

    [!TLDR]

    Substitute $u=x^2+y^2$; the equation becomes $\cos u\,du = 2\,dx$, which integrates to $\sin(x^2+y^2)=2x+c$.

    Concept

    This CBSE Class 12 differential equation is solved by a substitution that recognises the combination $x + y\frac{dy}{dx}$ as half of $\frac{d}{dx}(x^2+y^2)$.

    Solution

    Let $u = x^2 + y^2$. Then

    $\frac{du}{dx} = 2x + 2y\frac{dy}{dx} = 2\left(x + y\frac{dy}{dx}\right).$

    So $x + y\frac{dy}{dx} = \tfrac{1}{2}\frac{du}{dx}$. The given equation $x + y\frac{dy}{dx} = \sec(x^2+y^2)$ becomes

    $\frac{1}{2}\frac{du}{dx} = \sec u \;\Rightarrow\; \frac{du}{dx} = 2\sec u \;\Rightarrow\; \cos u\,du = 2\,dx.$

    Integrating both sides:

    $\sin u = 2x + c \;\Rightarrow\; \sin(x^2+y^2) = 2x + c.$

    [!ANSWER]

    (C) $\sin(x^2+y^2)=2x+c$

11. COMEDK 2025Set 2025-A1 markMCQ
    Q.Solution of the differential equation $y \frac{d y}{d x}+x=0$ represents a family of (A) Ellipse (B) Parabola (C) Circles (D) Hyperbola

    ›Reveal solutionSolution

    The differential equation $y \frac{dy}{dx} + x = 0$ is separable and integrates to $x^2 + y^2 = C$, which is the equation of a circle. The correct option is (C).

    The key is to recognize that this is a first-order separable differential equation. The form $y \frac{dy}{dx} = -x$ tells us we can move all $y$ terms to one side and all $x$ terms to the other, then integrate. The result will be a relation between $x$ and $y$ that describes a geometric curve. We don't need to solve for $y$ explicitly; the implicit equation reveals the family.

    1. Separate the variables. Start with $y \frac{dy}{dx} + x = 0$. Subtract $x$ from both sides:

    $y \frac{dy}{dx} = -x.$

    Multiply both sides by $dx$ (treating $dy/dx$ as a ratio for separation):

    $y \, dy = -x \, dx.$

    2. Integrate both sides. The left side is integrated with respect to $y$, the right with respect to $x$:

    $\int y \, dy = \int -x \, dx.$

    This gives:

    $\frac{y^2}{2} = -\frac{x^2}{2} + C_1,$

    where $C_1$ is an arbitrary constant of integration.

    3. Simplify to a standard form. Multiply through by 2:

    $y^2 = -x^2 + 2C_1.$

    Let $C = 2C_1$ (still an arbitrary constant). Then:

    $x^2 + y^2 = C.$

    4. Interpret the equation. The equation $x^2 + y^2 = C$ represents a circle centered at the origin $(0,0)$ with radius $\sqrt{C}$ (provided $C > 0$; if $C = 0$, it's just the point at the origin, and if $C < 0$, there are no real points). So the family of solutions is a family of concentric circles.

    Watch out

    A common mistake is to think the minus sign makes it a hyperbola ($x^2 - y^2 = \text{constant}$). But here both $x^2$ and $y^2$ have the same sign (positive) after rearranging, which is the signature of a circle (or ellipse with equal coefficients). Since the coefficients of $x^2$ and $y^2$ are both 1, it's specifically a circle, not a general ellipse.

    Tip

    You can also check by differentiating the implicit equation $x^2 + y^2 = C$: $2x + 2y \frac{dy}{dx} = 0$ simplifies to $y \frac{dy}{dx} + x = 0$, confirming the match.

    ✓Final answer

    The correct option is (C).

    ANSWER: C

12. COMEDK 2023Set 2023-M1 markMCQ
    Q.The general solution of $\left(\frac{d y}{d x}\right)^2=1-x^2-y^2+x^2 y^2$ is (A) $2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C$ (B) $\cos ^{-1} y=x \cos ^{-1} x$ (C) $\sin ^{-1} y=\frac{1}{2} \sin ^{-1} x+C$ (D) $2 \sin ^{-1} y=x \sqrt{1-y^2}+C$

    ›Reveal solutionSolution

    $1-x^2-y^2+x^2y^2=(1-x^2)(1-y^2)$; separating variables and integrating gives $2\sin^{-1}y=x\sqrt{1-x^2}+\sin^{-1}x+C$.

    Factor the right-hand side:

    $\left(\frac{dy}{dx}\right)^2=1-x^2-y^2+x^2y^2=(1-x^2)(1-y^2).$

    Taking the positive root and separating variables:

    $\frac{dy}{\sqrt{1-y^2}}=\sqrt{1-x^2}\,dx.$

    Integrate. The left side gives $\sin^{-1}y$. For the right side use $\int\sqrt{1-x^2}\,dx=\tfrac{x}{2}\sqrt{1-x^2}+\tfrac12\sin^{-1}x$:

    $\sin^{-1}y=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x+C'.$

    Multiplying through by 2:

    $2\sin^{-1}y=x\sqrt{1-x^2}+\sin^{-1}x+C.$

    ✓Final answer

    The correct option is (A) — $2 \sin ^{-1} y=x \sqrt{1-x^2}+\sin ^{-1} x+C$