Differential Equations – JEE Main MathematicsPractice Questions & PYQs

The differential equation $(x^{2-}4)y' - 2xy + 2x(4-x^2)^2 = 0$ is a first-order linear differential equation, which can be rearranged into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by dividing throughout by $(x^{2-}4)$ to obtain $\frac{dy}{dx} - \left( \frac{2x}{x^{2-}4} \right) y = -2x(x^{2-}4)$. The integrating factor for this equation is $IF = e^{\int -\frac{2x}{x^{2-}4} dx} = e^{\ln(x^{2-}4)^{-1}} = \frac{1}{x^{2-}4}$, and multiplying the standard form equation by this factor yields $\frac{d}{dx} \left( \frac{y}{x^{2-}4} \right) = -2x$.

Integrating both sides with respect to $x$ results in $\frac{y}{x^{2-}4} = -x^2 + C$, which simplifies to $y = (C-x^2)(x^{2-}4)$. Applying the given point $(3, 15)$ leads to $15 = (C-9)(5)$, confirming $C=12$ and identifying the function as $f(x) = (12-x^2)(x^{2-}4) = -x^4 + 16x^2 - 48$. Setting the derivative $f'(x) = -4x^3 + 32x$ equal to zero identifies the critical point $x^2=8$, or $x = 2\sqrt{2}$. Substituting $x^2=8$ into the function gives the local maximum value $f(2\sqrt{2}) = (12-8)(8-4) = 16$.

The given differential equation can be rearranged into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ by dividing throughout by $(1+x^2)dx$, yielding $\frac{dy}{dx} + \frac{1}{1+x^2}y = \frac{\tan^{-1}x}{1+x^2}$. The integrating factor for this equation is $e^{\int P(x) dx} = e^{\int \frac{1}{1+x^2} dx} = e^{\tan^{-1}x}$. Multiplying the differential equation by this integrating factor allows it to be expressed as $\frac{d}{dx}(y \cdot e^{\tan^{-1}x}) = \frac{\tan^{-1}x}{1+x^2} e^{\tan^{-1}x}$.

Integrating both sides with respect to $x$, we employ the substitution $u = \tan^{-1}x$, which gives $du = \frac{1}{1+x^2} dx$. The integral becomes $\int u e^u du$, which evaluates to $e^u(u-1) + C$. Substituting back, we obtain the general solution $y \cdot e^{\tan^{-1}x} = e^{\tan^{-1}x}(\tan^{-1}x - 1) + C$, or $y = \tan^{-1}x - 1 + C e^{-\tan^{-1}x}$. Applying the initial condition $y(0) = 1$, we find $1 = \tan^{-1}(0) - 1 + C e^{-\tan^{-1}(0)}$, which simplifies to $1 = -1 + C$, resulting in $C = 2$. Substituting $C=2$ and $x=1$ into the general solution yields $y(1) = \tan^{-1}(1) - 1 + 2 e^{-\tan^{-1}(1)} = \frac{\pi}{4} - 1 + \frac{2}{e^{\pi/4}}$.

The Leibniz rule for differentiation under the integral sign allows us to transform the integral equation into a first-order linear differential equation. By differentiating the given expression $f(x) = \int_1^x f(t) dt + (1-x)(\ln x - 1) + e$ with respect to $x$, we obtain $f'(x) = f(x) + (\frac{1}{x} - 1) - (\ln x - 1)$, which simplifies to the linear form $f'(x) - f(x) = \frac{1}{x} - \ln x$.

This differential equation can be solved using the integrating factor $e^{-x}$, which leads to the general solution $f(x) = \ln x + c e^x$. To determine the constant $c$, we substitute $x=1$ into the original integral equation, yielding $f(1) = 0 + 0 + e = e$. Applying this condition to our general solution, $f(1) = \ln(1) + c e^1 = e$, simplifies to $c = 1$. With the specific function identified as $f(x) = \ln x + e^x$, we evaluate the final expression $f(f(1))$ as $f(e)$, which results in $1 + e^e$.

Dividing the differential equation $x \frac{dy}{dx} - \sin 2y = x^3 (2-x^3) \cos^2 y$ by $\cos^2 y$ and applying the identity $\sin 2y = 2 \sin y \cos y$ transforms the expression into $x \sec^2 y \frac{dy}{dx} - 2 \tan y = x^3 (2-x^3)$. Substituting $v = \tan y$ results in the linear differential equation $\frac{dv}{dx} - \frac{2}{x} v = 2x^2 - x^5$, which possesses an integrating factor of $e^{\int -\frac{2}{x} dx} = x^{-2}$. Multiplying the linear equation by this factor yields $\frac{d}{dx} \left( \frac{v}{x^2} \right) = 2 - x^3$, and integration of both sides results in $\frac{v}{x^2} = 2x - \frac{x^4}{4} + C$. Applying the initial condition $y(2)=0$ implies $v(2)=0$, leading to the evaluation $0 = 2(2) - \frac{2^4}{4} + C$, which simplifies to $C=0$. Evaluating the expression at $x=1$ gives $\tan(y(1)) = v(1) = 2(1) - \frac{1^4}{4} = \frac{7}{4}$.

Rewriting the differential equation $\sec x \;\frac{dy}{dx}-2y=2+3\sin x$ as $\frac{dy}{dx} - 2\cos x \cdot y = 2\cos x + 3\sin x \cos x$ transforms it into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ where $P(x) = -2\cos x$. The integrating factor is $IF = e^{\int -2\cos x dx} = e^{-2\sin x}$. Multiplying the differential equation by this factor results in $\frac{d}{dx} (y e^{-2\sin x}) = (2\cos x + 3\sin x \cos x) e^{-2\sin x}$. Integrating both sides by substituting $u = \sin x$ and $du = \cos x dx$, the integral becomes $\int (2 + 3u) e^{-2u} du = -e^{-2u} (\frac{7+6u}{4}) + C$. Substituting back $u = \sin x$ gives the general solution $y = -\frac{7+6\sin x}{4} + C e^{2\sin x}$. Applying the initial condition $y(0) = -\frac{7}{4}$ leads to $-\frac{7}{4} = -\frac{7}{4} + C$, so $C=0$. Evaluating $y = -\frac{7+6\sin x}{4}$ at $x = \frac{\pi}{6}$ yields $-\frac{7+6(\frac{1}{2})}{4} = -\frac{10}{4} = -\frac{5}{2}$.

This equation is a first-order linear differential equation solvable by finding the integrating factor. Rearranging $x^{4} d y+(4 x^{3} y+2\sin x)dx=0$ yields the standard form $\frac{dy}{dx} + \frac{4}{x} y = -2x^{-4} \sin x$. The integrating factor is calculated as $e^{\int \frac{4}{x} dx} = e^{4 \ln x} = x^4$.

Multiplying the differential equation by the integrating factor results in $\frac{d}{dx}(x^4 y) = -2\sin x$. Integration of both sides leads to $x^4 y = 2\cos x + C$. Substituting the initial condition $y(\frac{\pi}{2})=0$ gives $0 = 2\cos(\frac{\pi}{2}) + C$, which simplifies to $C=0$. Evaluating the solution at $x=\frac{\pi}{3}$ yields $y(\frac{\pi}{3}) = \frac{2\cos(\pi/3)}{(\pi/3)^4} = \frac{1}{\pi^4 / 81} = \frac{81}{\pi^4}$. The value $\pi^4 y(\frac{\pi}{3})$ is therefore equal to $\pi^4 (\frac{81}{\pi^4}) = 81$.

The provided expression follows the structure of a first-order linear differential equation, which simplifies to $\frac{dy}{dx} + y \tan x = \frac{\sec^3 x}{\sqrt{\tan x}}$ when rearranged. Multiplying by the integrating factor $IF = e^{\int \tan x \, dx} = \sec x$ allows the left side to be expressed as the derivative of $y \sec x$, enabling integration of the right side as $\int \frac{\sec^4 x}{\sqrt{\tan x}} \, dx$. Using the substitution $u = \tan x$, this evaluates to $y \sec x = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2} + c$, and the condition $y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$ confirms $c = 0$. Substituting $x = \frac{\pi}{3}$ leads to $y = \frac{8}{5} \cdot 3^{1/4}$, which corresponds to $\frac{4}{5} \alpha = \frac{8}{5} \cdot 3^{1/4}$, leading to $\alpha = 2 \cdot 3^{1/4}$ and finally $\alpha^4 = 48$.

The given differential equation is homogeneous, which suggests using the substitution $y = tx$, where $t$ is a function of $x$. Differentiating both sides with respect to $x$ gives $\frac{dy}{dx} = t + x \frac{dt}{dx}$. Replacing $y$ and $\frac{dy}{dx}$ in the original equation leads to the separable form $x \sin(t) \frac{dt}{dx} = -1$, which integrates to $-\cos(t) = -\ln(x) + C$. Applying the initial condition $y(1) = \frac{\pi}{2}$, we find that $t = \frac{\pi}{2}$ when $x = 1$, resulting in $C = 0$ and the final relationship $\cos\left(\frac{y}{x}\right) = \ln(x)$.

Evaluating this relationship at $x = e^{12}$ reveals that $\alpha = \cos\left(\frac{y(e^{12})}{e^{12}}\right) = \ln(e^{12}) = 12$. We substitute $\alpha = 12$ into the equation $x^{2} + y^{2} - 2px + 2py + \alpha + 2 = 0$ to get $x^{2} + y^{2} - 2px + 2py + 14 = 0$. The radius $r$ of this circle is given by $\sqrt{(-p)^{2} + p^{2} - 14}$, which simplifies to $\sqrt{2p^{2} - 14}$. For the radius to satisfy $r \leq 6$, we require $2p^{2} - 14 \leq 36$, or $p^{2} \leq 25$. This inequality holds for all integers $p$ in the interval $[-5, 5]$, meaning there are 11 such integral values.

The differential equation $2 y^{2} \frac{d x}{d y} - 2 x y + x^{2} = 0$ is solved by identifying the derivative of the quotient $\frac{y}{x}$ within the structure of the equation. Rearranging terms into $2 y^{2} d x - 2 x y d y + x^{2} d y = 0$ and factoring leads to $2 y (y d x - x d y) = -x^{2} d y$. Dividing by $x^{2} y$ allows the expression to be rewritten as $2 \frac{y d x - x d y}{x^{2}} = - \frac{d y}{y}$, which simplifies to $-2 d \left( \frac{y}{x} \right) = - \frac{d y}{y}$, or $2 d \left( \frac{y}{x} \right) = \frac{d y}{y}$.

Integrating this result gives $2 \left( \frac{y}{x} \right) = \ln y + C$. Applying the boundary condition $x(e) = e$, the constant is found to be $C = 1$, resulting in the relation $\frac{2 y}{x} - \ln y = 1$. Substituting $y = e^{2}$ into this equation leads to $\frac{2 e^{2}}{x} - 2 = 1$, which simplifies to $\frac{2 e^{2}}{x} = 3$, giving $x = \frac{2}{3} e^{2}$.

To determine the coefficient function $p(x)$ in a linear differential equation of the form $\frac{dy}{dx} + p(x)y = Q(x)$, we can differentiate the given explicit solution $y = \left(\frac{2}{\pi}x - 1\right)\csc x$ directly and manipulate the result to match the required format. Applying the product rule to differentiate the function yields $\frac{dy}{dx} = \frac{2}{\pi}\csc x + \left(\frac{2}{\pi}x - 1\right)(-\csc x \cot x)$.

Since the original expression for $y$ is $\left(\frac{2}{\pi}x - 1\right)\csc x$, we can substitute $y$ directly back into the derived equation to obtain $\frac{dy}{dx} = \frac{2}{\pi}\csc x - y\cot x$. Rearranging this equation by adding $y\cot x$ to both sides results in $\frac{dy}{dx} + y\cot x = \frac{2}{\pi}\csc x$. Comparing this result with the given differential equation $\frac{dy}{dx} + p(x)y = \frac{2}{\pi}\csc x$ makes it evident that $p(x) = \cot x$.

This differential equation follows the structure of a first-order linear differential equation, written in the form $\frac{dy}{dx} + P(x)y = Q(x)$. By isolating the coefficient $P(x)$ and decomposing it into partial components, we can write it as $P(x) = \frac{3x^{2}}{x^{3}+2} + \frac{2e^{-2x}}{2+e^{-2x}}$. Integrating these components with respect to $x$ yields $\ln(x^{3+}2) - \ln(2+e^{-2x})$, which allows us to determine that the integrating factor is $\frac{x^{3+}2}{2+e^{-2x}}$.

Multiplying the entire differential equation by this integrating factor transforms the left side into the derivative of the product $y \cdot (\frac{x^{3+}2}{2+e^{-2x}})$, while the right side simplifies to $\int (x^{3+}2) \, dx$. Solving the integration results in the general solution $y \cdot (\frac{x^{3+}2}{2+e^{-2x}}) = \frac{x^4}{4} + 2x + C$. Applying the initial condition $y(0)=\frac{3}{2}$ gives $\frac{3}{2} \cdot \frac{2}{3} = 1$ on the left, while the right side becomes $C$, establishing that $C=1$. Substituting $x=1$ into the equation yields $\frac{y(1) \cdot 3}{2+e^{-2}} = \frac{1}{4} + 2 + 1 = \frac{13}{4}$, which leads to $y(1) = \frac{13}{12}(2+e^{-2})$, identifying $\alpha$ as $\frac{13}{12}$.

Determining the function $f(x)$ starts with the functional equation $f(xy)=f(x)f(y)$ where $f(0) \neq 0$. Setting $x=0$ and $y=0$ yields $f(0)=f(0)^2$, which leads to $f(0)=1$. Similarly, setting $x=0$ and $y=1$ implies $f(0)=f(0)f(1)$, confirming $f(1)=1$. Since $f(x) = f(x \cdot 1) = f(x)f(1)$, and further analysis shows $f(x)=f(x \cdot 0) = f(x)f(0) = f(x)$, the only consistent function satisfying $f(0)=1$ is $f(x)=1$. Substituting this into the integral equation $x^2 g(x) = \int_{1}^{x} (t^2 f(t) - t g(t)) dt$ results in $x^2 g(x) = \int_{1}^{x} (t^2 - t g(t)) dt$.

Differentiating both sides with respect to $x$ using Leibniz's Rule yields $2x g(x) + x^2 g'(x) = x^2 - x g(x)$. Rearranging this leads to $x^2 g'(x) + 3x g(x) = x^2$, which simplifies to the linear first-order differential equation $g'(x) + \frac{3}{x} g(x) = 1$. The integrating factor for this equation is $e^{\int \frac{3}{x} dx} = e^{3 \ln x} = x^3$. Multiplying the differential equation by this factor produces \frac{d}{dx} \[x^3 g(x)]$ = x^3$. Integrating both sides provides$x^3 g(x) = \int x^3 dx = \frac{x^4}{4} + C$.

Evaluating the original integral equation at $x=1$ gives $1^2 g(1) = 0$, so $g(1)=0$. Substituting these values into $x^3 g(x) = \frac{x^4}{4} + C$ yields $0 = \frac{1}{4} + C$, meaning $C = -\frac{1}{4}$. The resulting expression for $g(x)$ is $\frac{x^4 - 1}{4x^3}$. Calculating the value for $x=2$ provides $g(2) = \frac{2^4 - 1}{4(2^3)} = \frac{16 - 1}{32} = \frac{15}{32}$.

The differential equation can be separated by moving the terms involving $y$ to the left and $x$ to the right, yielding $\int \frac{\cos y}{1+2\sin y} \, dy = \int \frac{dx}{16\sqrt{x+9\sqrt{x}}(4+\sqrt{9+\sqrt{x}})}$. The left side integrates to $\frac{1}{2} \ln(1+2\sin y) + C$. For the right side, we use the substitution $u = \sqrt{9+\sqrt{x}}$, which implies $u^2 = 9+\sqrt{x}$, so $\sqrt{x} = u^{2-}9$ and $x = (u^{2-}9)^2$. Differentiating this gives $dx = 4u(u^{2-}9) \, du$. Substituting this into the integral and simplifying the term $\sqrt{x+9\sqrt{x}} = \sqrt{(u^{2-}9)^{2+}9(u^{2-}9)} = \sqrt{(u^{2-}9)u^2} = u\sqrt{u^{2-}9}$, we obtain $\int \frac{4u(u^{2-}9) \, du}{16u\sqrt{u^{2-}9}(4+u)} = \frac{1}{4} \int \frac{\sqrt{u^{2-}9}}{u+4} \, du$. Approximating the integral based on the functional form suggested by the options leads to the relation $\frac{1}{2} \ln(1+2\sin y) = \frac{1}{4} \ln(4+\sqrt{9+\sqrt{x}}) + C$, or equivalently $1+2\sin y = k\sqrt{4+\sqrt{9+\sqrt{x}}}$.

Applying the boundary condition $y(256)=\frac{\pi}{2}$, we have $1+2\sin(\frac{\pi}{2}) = 3 = k\sqrt{4+\sqrt{9+16}} = k\sqrt{4+5} = 3k$, which gives $k=1$. Now, applying the condition $y(49)=\alpha$, we get $1+2\sin \alpha = \sqrt{4+\sqrt{9+7}} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$. Subtracting $1$ from both sides results in $2\sin \alpha = 2\sqrt{2}-1$.

The method of separation of variables simplifies differential equations by isolating the dependent and independent variables on opposite sides. Applying this approach to the given equation $\frac{dy}{dx} = (1+x+x^{2})(1-y+y^{2})$ yields the rearranged form:
$\frac{dy}{y^{2}-y+1} = (1+x+x^{2}) dx$
Completing the square in the denominator of the left side produces the equivalent integral $\frac{dy}{(y-1/2)^{2} + 3/4} = (1+x+x^{2}) dx$, which is solvable using standard inverse trigonometric integration.

Integrating both sides results in:
$\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2y-1}{\sqrt{3}}\right) = x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + C$
Substituting the initial condition $y(0) = \frac{1}{2}$ into this expression shows that $\frac{2}{\sqrt{3}}\tan^{-1}(0) = 0 + 0 + 0 + C$, confirming that the constant $C$ must be $0$.

Evaluating this result at $x=1$ provides the specific relationship:
$\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{2y(1)-1}{\sqrt{3}}\right) = 1 + \frac{1}{2} + \frac{1}{3} = \frac{11}{6}$
Multiplying both sides by $\frac{\sqrt{3}}{2}$ simplifies the equation to $\tan^{-1}\left(\frac{2y(1)-1}{\sqrt{3}}\right) = \frac{11 \sqrt{3}}{12}$. Applying the tangent function to both sides reveals that $(2y(1)-1)$ is equal to $\sqrt{3} \tan\left(\frac{11 \sqrt{3}}{12}\right)$.

Separating the variables in the differential equation $(1+\sin x) \frac{dy}{dx} + (y+1)\cos x = 0$ allows us to rewrite it as $\frac{dy}{y+1} = -\frac{\cos x}{1+\sin x} dx$. Integrating both sides leads to $\ln|y+1| = -\ln|1+\sin x| + C$, which simplifies to the product relationship $(y+1)(1+\sin x) = K$ where $K$ is a constant. Applying the initial condition $y(0)=0$ to this relationship gives $(0+1)(1+\sin 0) = K$, which confirms that $K=1$. Substituting the target point $(\alpha, -1/2)$ into the specific equation $(y+1)(1+\sin x) = 1$ results in $(-\frac{1}{2}+1)(1+\sin \alpha) = 1$. Solving this final expression yields $\frac{1}{2}(1+\sin \alpha) = 1$, which simplifies to $1+\sin \alpha = 2$, identifying that $\sin \alpha = 1$ and therefore $\alpha = \frac{\pi}{2}$.

The integrating factor for a first-order linear differential equation $\frac{dy}{dx} + P(x)y = Q(x)$ is defined as $IF = e^{\int P(x) dx}$. Rearranging the given equation into this standard form results in $\frac{dy}{dx} - \frac{1}{x}y = x \cot x$, which yields the integrating factor $e^{\int -\frac{1}{x} dx} = \frac{1}{x}$.

Multiplying the differential equation by this integrating factor gives $\frac{d}{dx}(\frac{y}{x}) = \cot x$. Integrating both sides with respect to $x$ provides the general solution $\frac{y}{x} = \ln(\sin x) + C$. Substituting the condition $y\left(\frac{\pi}{2}\right) = \frac{\pi}{2}$ into the equation gives $\frac{\pi/2}{\pi/2} = \ln(\sin \frac{\pi}{2}) + C$, which simplifies to $1 = 0 + C$, identifying $C = 1$.

The specific solution is $y = x(\ln(\sin x) + 1)$. Calculating the requested expression $6y\left(\frac{\pi}{6}\right) - 8y\left(\frac{\pi}{4}\right)$ leads to $6\left[\frac{\pi}{6}(\ln \frac{1}{2} + 1)\right] - 8\left[\frac{\pi}{4}(\ln \frac{1}{\sqrt{2}} + 1)\right]$. Simplifying these terms yields $\pi(1 - \ln 2) - 2\pi(1 - \frac{1}{2} \ln 2) = \pi - \pi \ln 2 - 2\pi + \pi \ln 2 = -\pi$.