Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Rearranging the differential equation $\sqrt{4-x^2} \frac{dy}{dx} = \left( \left(\sin^{-1}\left(\frac{x}{2}\right)\right)^2 - y \right) \sin^{-1}\left(\frac{x}{2}\right)$ into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ results in:

$\frac{dy}{dx} + \left( \frac{\sin^{-1}(x/2)}{\sqrt{4-x^2}} \right) y = \frac{(\sin^{-1}(x/2))^3}{\sqrt{4-x^2}}$

By substituting $t = \sin^{-1}(x/2)$, which implies $dt = \frac{dx}{\sqrt{4-x^2}}$, the integrating factor becomes $e^{\int t \, dt} = e^{t^2/2}$. Multiplying the linear equation by this factor and integrating both sides yields $y e^{t^2/2} = \int t^3 e^{t^2/2} \, dt$. Using integration by parts with the substitution $u = t^2/2$, the integral evaluates to $2(u - 1)e^u + C$, simplifying the general solution to $y = (\sin^{-1}(x/2))^2 - 2 + C e^{-\frac{(\sin^{-1}(x/2))^2}{2}}$. Applying the condition $y(2) = \frac{\pi^{2-}8}{4}$ requires $\sin^{-1}(1) = \pi/2$, which forces the constant $C$ to be $0$ and yields the particular solution $y = (\sin^{-1}(x/2))^2 - 2$. Evaluating this expression at $x=0$ gives $y(0) = -2$, so $y^2(0) = 4$.

Separating variables is the primary technique applied to this differential equation because it isolates $y$ terms on one side and $x$ terms on the other. Beginning with the equation $2(3+y) e^{2x} dx - (7+e^{2x}) dy = 0$, we rearrange the terms to obtain:

$\frac{dy}{3+y} = \frac{2e^{2x}}{7+e^{2x}} dx$

Integrating both sides yields $\ln(3+y) = \ln(7+e^{2x}) + C$, where $C$ is an integration constant. Through exponentiation, this relationship simplifies to $3+y = A(7+e^{2x})$, with $A$ representing a constant.

Applying the point $(0,5)$ to determine $A$ involves substituting $x=0$ and $y=5$ into the equation, which results in $3+5 = A(7+e^0)$. This equation simplifies to $8 = 8A$, establishing that $A=1$. Consequently, the equation for the curve is $y = e^{2x} + 4$. Evaluation at the point $(\ln 2, k)$ requires substituting $x = \ln 2$, which gives $k = e^{2 \ln 2} + 4$. Since $e^{2 \ln 2}$ simplifies to $4$, we find $k=8$.

The structure of this differential equation, involving a ratio $x/y$, suggests using the substitution $x=ty$, which leads to $\frac{dx}{dy} = t + y\frac{dt}{dy}$. Substituting this into the equation $y = \left(x-y \frac{dx}{dy}\right) \sin \left(\frac{x}{y}\right)$ simplifies the expression to $1 = -y \sin(t) \frac{dt}{dy}$ after dividing by $y$. This form is separable, allowing us to integrate $\frac{dy}{y} = -\sin(t) dt$ to obtain $\ln y = \cos(t) + C$, which results in $\ln y = \cos \left(\frac{x}{y}\right) + C$ after substituting $t = \frac{x}{y}$ back into the equation.

Given the condition $x(1)=\frac{\pi}{2}$, we find $\ln(1) = \cos(\frac{\pi}{2}) + C$, which confirms that $C=0$, simplifying the relation to $\cos \left(\frac{x}{y}\right) = \ln y$. Solving for $x$ at $y=2$ gives $x(2) = 2 \cos^{-1}(\ln 2)$. Using the trigonometric double-angle formula $\cos(2\theta) = 2\cos^2 \theta - 1$ where $\theta = \cos^{-1}(\ln 2)$, we evaluate $\cos(x(2))$ to obtain $2 (\log_e 2)^2 - 1$.

This problem centers on a first-order linear differential equation, which takes the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. To isolate the dependent variable, we calculate the integrating factor, defined as $e^{\int P(x) \, dx}$. Given the term $\tan x$, the integral $\int \tan x \, dx$ results in $\ln(\sec x)$, making the integrating factor $\sec x$. Multiplying the entire equation by $\sec x$ allows the left side to be expressed as the derivative of the product $y \sec x$, setting the stage for direct integration of the right side.

The integral of the right side, $\int \frac{(2+\sec x) \sec x}{(1+2\sec x)^2} \, dx$, becomes manageable through the substitution $t = \tan(x/2)$, where $dx = \frac{2}{1+t^2} \, dt$. Using the identities $\sec x = \frac{1+t^2}{1-t^2}$ and $\tan(x/2) = t$, the integral simplifies to $2 \int \frac{3-t^2}{(3+t^2)^2} \, dt$. This expression resolves to $\frac{2t}{t^{2+}3} + C$, yielding the general solution $y \sec x = \frac{2 \tan(x/2)}{\tan^2(x/2)+3} + C$.

We determine the constant of integration by substituting the initial condition $f\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{10}$. Since $\sec\left(\frac{\pi}{3}\right) = 2$ and $\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$, we solve $\left(\frac{\sqrt{3}}{10}\right) \cdot 2 = \frac{2(1/\sqrt{3})}{(1/3)+3} + C$, which simplifies to $\frac{2\sqrt{3}}{10} = \frac{2\sqrt{3}}{10} + C$, confirming that $C=0$. With the specific solution $y = \frac{2 \tan(x/2)}{\sec x (\tan^2(x/2)+3)}$ now established, substituting $x = \frac{\pi}{4}$ requires calculating $\tan(\pi/8) = \sqrt{2}-1$ and $\sec(\pi/4) = \sqrt{2}$. Plugging these values into the function yields $\frac{2(\sqrt{2}-1)}{\sqrt{2}((\sqrt{2}-1)^2 + 3)}$, which simplifies to $\frac{4-\sqrt{2}}{14}$.

This differential equation follows the standard linear form $\frac{dy}{dx}+P(x)y=Q(x)$, where $P(x)=\frac{x}{x^{2-}1}$. Determining the integrating factor involves calculating $e^{\int \frac{x}{x^{2-}1}dx}$, which simplifies to $e^{\frac{1}{2}\ln(1-x^2)}$, yielding $\sqrt{1-x^2}$ for the given interval. Multiplying the entire equation by this factor transforms it into the form $\frac{d}{dx}(y\sqrt{1-x^2}) = x^{6+}4x$. Integrating both sides results in $y\sqrt{1-x^2} = \frac{x^7}{7}+2x^{2+}C$, and since $f(0)=0$, the constant $C$ must be zero, giving $f(x)=\frac{x^7}{7\sqrt{1-x^2}}+\frac{2x^2}{\sqrt{1-x^2}}$.

Evaluating the definite integral $\int_{-\frac{1}{2}}^{\frac{1}{2}} f(x)dx$ simplifies because the first term contains an odd power of $x$, rendering its integral zero over the symmetric interval $[-\frac{1}{2}, \frac{1}{2}]$. We are left to compute $\int_{-\frac{1}{2}}^{\frac{1}{2}} \frac{2x^2}{\sqrt{1-x^2}}dx$. Substituting $x=\sin\theta$ and $dx=\cos\theta d\theta$ transforms the limits to $[-\frac{\pi}{6}, \frac{\pi}{6}]$ and the integrand to $2\sin^2\theta$. Calculating $2\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} \sin^2\theta d\theta$ is equivalent to $\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}} (1-\cos 2\theta)d\theta$, which evaluates to $\frac{\pi}{3}-\frac{\sqrt{3}}{2}$. Multiplying this result by six produces $2\pi-3\sqrt{3}$, revealing that $\alpha=3\sqrt{3}$ and therefore $\alpha^2=27$.

The differential equation fits the linear standard form $\frac{dy}{dx} - \left(\frac{2x}{1+x^2}\right) y = (1+x^2) \cos x$ after dividing both sides by $(1+x^2)$. An integrating factor is computed as $e^{-\int \frac{2x}{1+x^2} dx} = \frac{1}{1+x^2}$, which simplifies the left side of the equation to the derivative $\frac{d}{dx} \left( \frac{y}{1+x^2} \right)$. Equating this to the right side and integrating leads to $\frac{y}{1+x^2} = \sin x + c$, which satisfies the initial condition $y(0)=1$ when $c=1$. This yields the function $y = (1+\sin x)(1+x^2)$, and evaluating the integral over the symmetric interval from $-3$ to $3$ leverages the symmetry of the expression. The odd component, $\sin x(1+x^2)$, integrates to zero over this interval, leaving twice the integral of the even part $(1+x^2)$ from $0$ to $3$, which evaluates to $2 \left[ x + \frac{x^3}{3} \right]_0^3 = 2(3+9) = 24$.

A first-order linear differential equation is solved by arranging it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Rearranging the terms of the original equation yields $\frac{dy}{dx} + \frac{x}{1+x^2} y = \frac{5x^2}{\sqrt{1+x^2}}$, which identifies the functions $P(x) = \frac{x}{1+x^2}$ and $Q(x) = \frac{5x^2}{\sqrt{1+x^2}}$. The integrating factor, defined as $e^{\int P(x) dx}$, simplifies to $e^{\int \frac{x}{1+x^2} dx} = \sqrt{1+x^2}$.

Multiplying the differential equation by this integrating factor produces \frac{d}{dx} \[y \sqrt{1+x^2}]$ = 5x^2$. Integrating both sides with respect to$x$results in$y \sqrt{1+x^2} = \frac{5x^3}{3} + C$. Using the initial condition$y(0)=0$, we determine that$C=0$. Evaluating the expression at$x = \sqrt{3}$gives$y(\sqrt{3}) = \frac{5(\sqrt{3})^3}{3 \sqrt{1+(\sqrt{3})^2}}$, which simplifies to$\frac{5\sqrt{3}}{2}$.

Begin by evaluating the functional equation $f(x+y)=f(x) f'(y)+f'(x) f(y)$ at $x=0$ and $y=0$. Since $f(0)=1$, this substitution yields $f(0) = f(0)f'(0) + f'(0)f(0)$, which simplifies to $1 = 2f'(0)$ and reveals that $f'(0) = \frac{1}{2}$. By setting $y=0$ in the original functional equation, we obtain $f(x) = f(x)f'(0) + f'(x)f(0)$, or $f(x) = \frac{1}{2}f(x) + f'(x)$, which reduces to the differential equation $f'(x) = \frac{1}{2}f(x)$.

Integrating the ratio $\frac{f'(x)}{f(x)} = \frac{1}{2}$ leads to the expression $\ln f(x) = \frac{x}{2} + C$. The initial condition $f(0)=1$ implies $\ln(1) = 0 + C$, confirming that $C=0$ and $\ln f(x) = \frac{x}{2}$. Consequently, the required summation $\sum_{n=1}^{100} \ln f(n)$ simplifies to $\sum_{n=1}^{100} \frac{n}{2}$. Using the formula for the sum of the first $n$ integers, the calculation becomes $\frac{1}{2} \cdot \frac{100(101)}{2}$, which equals $\frac{5050}{2}$, or 2525.

Treating $x$ as the dependent variable reveals this as a first-order linear differential equation. Dividing the original equation by $\frac{dy}{dx}$ and rearranging the terms leads to $(1+y^2) \frac{dx}{dy} + x = 2 e^{\tan^{-1} y}$. Dividing through by $1+y^2$ puts the expression into the standard linear form:

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

The integrating factor is computed as $e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$. Multiplying both sides by this factor transforms the left side into the derivative of the product $x \cdot e^{\tan^{-1} y}$, while the right side becomes $\frac{2 (e^{\tan^{-1} y})^2}{1+y^2}$. Integrating this yields $x \cdot e^{\tan^{-1} y} = e^{2 \tan^{-1} y} + c$.

Substituting the condition $f(0)=1$, which implies $x=1$ when $y=0$, confirms that $1 \cdot e^0 = e^0 + c$, so $c=0$. This simplifies the relationship to $x = e^{\tan^{-1} y}$. Consequently, evaluating $f\left(\frac{1}{\sqrt{3}}\right)$ gives $e^{\tan^{-1} \left(\frac{1}{\sqrt{3}}\right)}$, which evaluates to $e^{\frac{\pi}{6}}$.

Differentiating both sides of the integral equation with respect to $x$ provides a clear path toward determining $f(x)$. Applying the Fundamental Theorem of Calculus to the integral term and utilizing the product rule for the expression $2(x+2)^2 f(x)$, we obtain the differentiated equation $4(x+2) f(x) + 2(x+2)^2 f'(x) - 6(x+2) = 10(x+2) f(x)$.

By rearranging terms and dividing by $2(x+2)$, we simplify the expression to the linear differential equation $f'(x) - \frac{3}{x+2} f(x) = \frac{3}{x+2}$. Using the integrating factor $(x+2)^{-3}$, we can integrate both sides to find that $f(x) = C(x+2)^3 - 1$, where $C$ is a constant.

To solve for the constant $C$, we evaluate the original integral equation at $x=0$, which gives $2(2)^2 f(0) - 3(2)^2 = 0$, establishing that $f(0) = \frac{3}{2}$. Substituting $x=0$ and $f(0) = \frac{3}{2}$ into our expression for $f(x)$, we find $\frac{3}{2} = 8C - 1$, which simplifies to $C = \frac{5}{16}$. With $f(x) = \frac{5}{16}(x+2)^3 - 1$ defined, we evaluate $f(2)$ by calculating $\frac{5}{16}(4)^3 - 1$, leading to the result 19.

The Fundamental Theorem of Calculus allows us to transform integral equations into differential ones by differentiating with respect to the variable upper limit. Applying this to both sides of the equation $\int_{0}^{x} t f(t) dt = x^2 f(x)$ yields $x f(x)$ on the left, while the product rule applied to the right side gives $2x f(x) + x^2 f'(x)$. Equating these results leads to $x f(x) = 2x f(x) + x^2 f'(x)$, which simplifies to $x^2 f'(x) = -x f(x)$.

Dividing both sides by $x^2 f(x)$ leaves the separable form $\frac{f'(x)}{f(x)} = -\frac{1}{x}$. Integration produces $\ln f(x) = -\ln x + C$, which can be expressed as $f(x) = \frac{k}{x}$ for some constant $k$. Substituting the provided condition $f(2) = 3$ into this expression gives $3 = \frac{k}{2}$, identifying $k=6$. With the function fully determined as $f(x) = \frac{6}{x}$, calculating $f(6)$ results in $f(6) = 1$.

The differential equation $(7 x^4 \cot y - e^x \csc y) \frac{dx}{dy} = x^5$ can be rewritten by taking the reciprocal as $\frac{dy}{dx} = \frac{x^5 \sin y}{7 x^4 \cos y - e^x}$ after substituting $\cot y = \frac{\cos y}{\sin y}$ and $\csc y = \frac{1}{\sin y}$. Multiplying both sides by $\sin y$ yields $\sin y \frac{dy}{dx} = \frac{7}{x} \cos y - \frac{e^x}{x^5}$, which transforms into the linear form $\frac{dv}{dx} + \frac{7}{x} v = \frac{e^x}{x^5}$ by substituting $v = \cos y$ and $\frac{dv}{dx} = -\sin y \frac{dy}{dx}$.

Multiplying this linear equation by the integrating factor $e^{\int (7/x) dx} = x^7$ gives $\frac{d}{dx} (x^7 v) = x^2 e^x$. Integrating both sides leads to $x^7 v = \int x^2 e^x dx = e^x(x^2 - 2x + 2) + C$. Using the initial condition $v(1) = \cos(\frac{\pi}{2}) = 0$, we find $0 = e^1(1^2 - 2(1) + 2) + C$, which implies $C = -e$. The final expression for $\cos y$ is $v = \frac{e^x(x^2 - 2x + 2) - e}{x^7}$, and evaluating this at $x=2$ gives $\frac{e^2(2^2 - 2(2) + 2) - e}{2^7} = \frac{2e^2 - e}{128}$.

This problem is a first-order linear differential equation that can be streamlined using the trigonometric identity $1+\tan^2 x = \sec^2 x$. The original expression $\frac{d y}{d x}+3(\tan^2 x) y+3 y=\sec^2 x$ simplifies by factoring $3y$ out of the second and third terms to become $\frac{d y}{d x}+3 \sec^2 x \cdot y=\sec^2 x$. The integrating factor is obtained by exponentiating the integral of the coefficient of $y$, which results in $I.F. = e^{\int 3 \sec^2 x \, dx} = e^{3 \tan x}$.

Multiplying the differential equation by this integrating factor turns the left side into the derivative of the product $y \cdot e^{3 \tan x}$, such that $\frac{d}{d x}(y \cdot e^{3 \tan x}) = \sec^2 x \cdot e^{3 \tan x}$. Integrating both sides with respect to $x$ yields $y \cdot e^{3 \tan x} = \frac{1}{3} e^{3 \tan x} + C$. Applying the initial condition $y(0) = \frac{1}{3} + e^3$ determines that the constant $C$ equals $e^3$. Substituting $x = \frac{\pi}{4}$ into $y \cdot e^{3 \tan x} = \frac{e^{3 \tan x}}{3} + e^3$ gives $y \cdot e^3 = \frac{e^3}{3} + e^3$, simplifying to $y = \frac{4}{3}$.

This problem involves a first-order linear differential equation that can be standardly expressed in the form $\frac{dy}{dx} + P(x)y = Q(x)$. To reach this form, rewrite the given equation by dividing throughout by $\cos x (\ln(\cos x))^2 dx$, which isolates the derivative term and reveals $P(x) = -\frac{3 \tan x}{\ln(\cos x)}$ and $Q(x) = -\frac{\tan x}{(\ln(\cos x))^2}$. The integrating factor is found using $e^{\int P(x) \, dx}$, where the substitution $t = \ln(\cos x)$ simplifies the integral to $3 \int \frac{1}{t} \, dt$, resulting in an integrating factor of $(\ln(\cos x))^3$.

Multiplying the differential equation by this integrating factor reduces the left side to the derivative of the product $y(\ln(\cos x))^3$, setting up the integration \frac{d}{dx} \[y(\ln(\cos x))^3]$ = -\tan x \ln(\cos x)$. Integrating both sides with respect to$x$and applying the same substitution$u = \ln(\cos x)$yields$y(\ln(\cos x))^3 = \int u , du = \frac{(\ln(\cos x))^2}{2} + C$. Using the initial condition$y(\pi/4) = -1/\ln 2$, we find$C = 0$, leading to the solution$y = \frac{1}{2 \ln(\cos x)}$. Finally, substituting$x = \pi/6$into this result gives$y = \frac{1}{2 \ln(\sqrt{3}/2)}$, which evaluates to$\frac{1}{\ln 3 - \ln 4}$.

This differential equation belongs to the linear class $\frac{dx}{dy} + P(y)x = Q(y)$, which is solved by finding an integrating factor to simplify the derivative into a product rule form. By rearranging the terms into $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$, the integrating factor is determined as $e^{\int \frac{1}{y^2} dy}$, which evaluates to $e^{-\frac{1}{y}}$. Multiplying the equation by this factor allows both sides to be integrated, yielding $x e^{-\frac{1}{y}} = \int \frac{1}{y^3} e^{-\frac{1}{y}} dy + C$.

Applying the substitution $t = -\frac{1}{y}$, where $dt = \frac{1}{y^2} dy$, transforms the integral into the expression $-\int t e^t dt$, which evaluates to $-e^t(t - 1) + C$. Substituting $t$ back gives the general solution $x e^{-\frac{1}{y}} = -e^{-\frac{1}{y}}(-\frac{1}{y} - 1) + C$, which simplifies to $x = \frac{1}{y} + 1 + C e^{\frac{1}{y}}$. Using the initial condition $x(1) = 1$ leads to $1 = 1 + 1 + C e$, resulting in $C = -\frac{1}{e}$. Substituting $y = \frac{1}{2}$ into the specific solution $x = \frac{1}{y} + 1 - e^{\frac{1}{y} - 1}$ provides $x = 2 + 1 - e^{2-1}$, resulting in $3-e$.

A first-order linear differential equation is solved by rearranging it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing the original equation $2 \cos x \frac{dy}{dx} = \sin 2x - 4y \sin x$ by $2 \cos x$ yields $\frac{dy}{dx} + 2y \tan x = \sin x$. The integrating factor is calculated as $e^{\int 2 \tan x \, dx}$, which simplifies to $\sec^2 x$. Multiplying the rearranged equation by this factor transforms the left side into the derivative of a product, giving $y \sec^2 x = \int \sin x \sec^2 x \, dx$. The integral on the right is equivalent to $\int \frac{\sin x}{\cos^2 x} \, dx$, which evaluates to $\sec x + C$.