Q.Solve the following differential equation: $y = \sqrt{a^2 - x^2}$ $x \in (-a, a)$ ; $x + y \frac{dy}{dx} = 0 (y \neq 0)$

Showing the 12 most recent of 13 on this concept.

1. 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 …

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2. 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. $\sin^{-1}y=-\sin^{-1}x+c$. …

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3. 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. …

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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 $x=0$, $y=\frac12$: $\sin^{-1}(0) + \sin^{-1}\!\left(\frac12\right) = C$ …

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5. 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|. …

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6. 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 $\int\sqrt{1-x^2}\,dx=\tfrac{x}{2}\sqrt{1-x^2}+\tfrac12\sin^{-1}x$: …

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7. 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 $\cos y$:

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

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8. COMEDK 2024Set 2024-A1 markMCQ
   Q.

   [!FORMULA] $\text { The general solution of } \frac{d y}{d x}=\sin ^{-1} x \text { is }$

   (A) $y=x \sin ^{-1} x+\sqrt{1-x^2}+C$ (B) $y=x \sin ^{-1} x-\sqrt{1-x^2}+C$ (C) $y=-x \sin ^{-1} x+\sqrt{1-x^2}+C$ (D) $y=-x \sin ^{-1} x-\sqrt{1-x^2}+C$

   ›Reveal solutionSolution

   The differential equation is solved by direct integration: $y = \int \sin^{-1} x \, dx$. Using integration by parts, the result is $y = x \sin^{-1} x + \sqrt{1-x^2} + C$, which matches option (A).

   Concept & Intuition

   This is a simple first-order differential equation of the form $\frac{dy}{dx} = f(x)$. The general solution is just the indefinite integral of $f(x)$ with respect to $x$, plus a constant of integration. The only challenge here is integrating $\sin^{-1} x$, which is an inverse trigonometric function. Integration by parts is the natural choice because we know the derivative of $\sin^{-1} x$ is $\frac{1}{\sqrt{1-x^2}}$, and that derivative will simplify the integral.

   Step-by-step solution

   1. Set up the integration Since $\frac{dy}{dx} = \sin^{-1} x$, we integrate both sides:

   $y = \int \sin^{-1} x \, dx + C.$

   The constant $C$ will be determined by initial conditions (if any), but here we want the general solution.

   2. Apply integration by parts

      Recall the formula: $\int u \, dv = uv - \int v \, du$.

      Choose $u = \sin^{-1} x$ and $dv = dx$.

      Then $du = \frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.

   3. Carry out the substitution

   $\int \sin^{-1} x \, dx = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1-x^2}} \, dx.$

   4. Simplify the remaining integral The integral $\int \frac{x}{\sqrt{1-x^2}} \, dx$ is a standard form. Let $t = 1 - x^2$, so $dt = -2x \, dx$, hence $x \, dx = -\frac{1}{2} dt$. Then:

   $\int \frac{x}{\sqrt{1-x^2}} \, dx = \int \frac{1}{\sqrt{t}} \left(-\frac{1}{2}\right) dt = -\frac{1}{2} \int t^{-1/2} dt = -\frac{1}{2} \cdot 2 t^{1/2} = -\sqrt{1-x^2}.$

   5. Combine results Substituting back: …

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9. 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.

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

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10. 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: $$$$

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11. 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:

    $|\cos y| = \frac{k}{|x|}$ …

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12. 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

    $\log y = \frac{c_1}{\sin x}.$ …

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