Differential Equations – JEE Main MathematicsPractice Questions & PYQs

The given equation $y^{\frac{1}{4}}+y^{-\frac{1}{4}}=2x$ can be processed by squaring both sides, resulting in $(y^{\frac{1}{4}}+y^{-\frac{1}{4}})^2 = 4x^2$. Using the algebraic identity $a-b = \sqrt{(a+b)^2 - 4ab}$, we can express the difference as $y^{\frac{1}{4}}-y^{-\frac{1}{4}} = \sqrt{(2x)^2 - 4} = 2\sqrt{x^{2-}1}$. Differentiating the original relation $y^{\frac{1}{4}}+y^{-\frac{1}{4}}=2x$ with respect to $x$ yields $\frac{1}{4}y^{-\frac{3}{4}}y' - \frac{1}{4}y^{-\frac{5}{4}}y' = 2$. Multiplying this entire equation by $4y$ simplifies it to $(y^{\frac{1}{4}}-y^{-\frac{1}{4}})y' = 8y$. Substituting our derived expression for the difference of powers gives $2\sqrt{x^{2-}1}y' = 8y$, which simplifies to $\sqrt{x^{2-}1}y' = 4y$. Squaring both sides results in $(x^{2-}1)(y')^2 = 16y^2$. Differentiating this expression with respect to $x$ provides $(x^{2-}1) \cdot 2y' y'' + 2x(y')^2 = 32y y'$. Dividing the entire equation by $2y'$ leaves $(x^{2-}1)y'' + x y' = 16y$, or $(x^{2-}1)y'' + x y' - 16y = 0$. Comparing this result with the given differential equation $(x^{2-}1) \frac{d^2y}{dx^2} + \alpha x \frac{dy}{dx} + \beta y = 0$, we identify the constants as $\alpha = 1$ and $\beta = -16$. Calculating the required value, we find $|\alpha - \beta| = |1 - (-16)| = 17$.

Simplifying this differential equation starts by recognizing that $2^x$ is a common factor in both the numerator and the denominator, which allows for cancelling that term to rewrite the expression as $\frac{dy}{dx} = \frac{y+2^y}{1+2^y \ln 2}$. This structure immediately facilitates separating the variables, leading to the integral form $\int \frac{1+2^y \ln 2}{y+2^y} dy = \int dx$. Because the numerator is precisely the derivative of the denominator with respect to $y$, the integral simplifies directly to $\ln |y+2^y| = x + C$. Applying the initial condition $y(0)=0$ results in $\ln |0+2^0| = 0 + C$, which identifies the constant $C$ as $0$. With the resulting equation $x = \ln(y+2^y)$, substituting $y=1$ gives $x = \ln(1+2^1) = \ln 3$. Since $e^1 \approx 2.718$ and $e^2 \approx 7.389$, the value of $\ln 3$ must reside in the interval $(1, 2)$.

This differential equation is a classic example of a Bernoulli equation, which we can solve by transforming it into a linear first-order equation. By rewriting the original equation as $\frac{dy}{dx} = (y+1)^2 e^{x^2/2} - x(y+1)$ and dividing both sides by $(y+1)^2$, we obtain the following relation:
$\frac{1}{(y+1)^2} \frac{dy}{dx} + \frac{x}{y+1} = e^{x^2/2}$
Using the substitution $v = \frac{1}{y+1}$, we find its derivative $\frac{dv}{dx} = -\frac{1}{(y+1)^2} \frac{dy}{dx}$, which allows us to convert the equation into the linear form $\frac{dv}{dx} - xv = -e^{x^2/2}$.

To solve this linear equation, we calculate the integrating factor $e^{\int -x dx} = e^{-x^2/2}$. Multiplying the entire equation by this factor yields $\frac{d}{dx} \left( v e^{-x^2/2} \right) = -1$. Integrating both sides with respect to $x$ gives $v e^{-x^2/2} = -x + C$, or equivalently, $\frac{1}{y+1} e^{-x^2/2} = -x + C$. Applying the initial condition $y(2)=0$, we substitute $x=2$ into this equation to find the constant: $\frac{1}{1} e^{-2} = -2 + C$, which simplifies to $C = 2 + e^{-2}$.

Now we evaluate the state of the system at $x=1$. Substituting $x=1$ and our constant $C$ back into the solution gives $\frac{1}{y+1} e^{-1/2} = -1 + 2 + e^{-2} = 1 + e^{-2}$. Solving for the term $(y+1)$ yields $y+1 = \frac{e^{-1/2}}{1 + e^{-2}}$. Finally, we substitute this value and $x=1$ back into the original differential equation $\frac{dy}{dx} = (y+1) [ (y+1) e^{x^2/2} - x ]$:
$y'(1) = \frac{e^{-1/2}}{1+e^{-2}} \left( \frac{e^{-1/2}}{1+e^{-2}} e^{1/2} - 1 \right) = \frac{e^{-1/2}}{1+e^{-2}} \left( \frac{1}{1+e^{-2}} - 1 \right)$
Simplifying the expression within the parentheses as $\frac{1-(1+e^{-2})}{1+e^{-2}} = \frac{-e^{-2}}{1+e^{-2}}$, we arrive at:
$y'(1) = \frac{-e^{-5/2}}{(1+e^{-2})^2} = \frac{-e^{-5/2}}{\frac{(e^{2+}1)^2}{e^4}} = \frac{-e^{3/2}}{(e^{2+}1)^2}$

Dividing the differential equation $x dy - y dx = \sqrt{x^{2} - y^{2}} dx$ by $x^{2}$ enables us to rewrite the expression as $\frac{x dy - y dx}{x^{2}} = \frac{\sqrt{x^{2} - y^{2}}}{x^{2}} dx$. By substituting $v = \frac{y}{x}$, we recognize that $dv = \frac{x dy - y dx}{x^{2}}$ and the right side simplifies to $\frac{\sqrt{1 - v^{2}}}{x} dx$. This yields the separable form $\frac{dv}{\sqrt{1 - v^{2}}} = \frac{dx}{x}$, which integrates directly to $\sin^{-1}(v) = \ln(x) + C$ because $x \ge 1$. Applying the initial condition $y(1) = 0$, where $v = 0$ at $x = 1$, we find $\sin^{-1}(0) = \ln(1) + C$, indicating that $C = 0$. Consequently, the solution is $y = x \sin(\ln x)$.

To determine the area bounded by this curve, the x-axis, and the lines $x=1$ and $x=e^{\pi}$, we calculate the definite integral $\int_{1}^{e^{\pi}} x \sin(\ln x) dx$. By performing the substitution $t = \ln x$, we obtain $dx = e^{t} dt$, transforming the integral into $\int_{0}^{\pi} e^{t} \sin(t) e^{t} dt$, which simplifies to $\int_{0}^{\pi} e^{2t} \sin(t) dt$. Utilizing the integral formula $\int e^{at} \sin(bt) dt = \frac{e^{at}}{a^{2} + b^{2}} (a \sin(bt) - b \cos(bt))$ with $a=2$ and $b=1$, we evaluate the expression over the limits from $0$ to $\pi$. This calculation results in $\left[ \frac{e^{2t}}{5} (2 \sin t - \cos t) \right]_{0}^{\pi}$, which expands to $\frac{e^{2\pi}}{5}(0 - (-1)) - \frac{1}{5}(0 - 1) = \frac{e^{2\pi}}{5} + \frac{1}{5}$. Since the area is given in the form $\alpha e^{2\pi} + \beta$, we identify $\alpha = \frac{1}{5}$ and $\beta = \frac{1}{5}$, making the value of $10(\alpha + \beta)$ equal to $10\left(\frac{1}{5} + \frac{1}{5}\right)$, which results in $4$.

The principle of linearity in differential equations allows us to solve first-order problems by identifying an integrating factor that simplifies the expression into a product rule derivative. Transforming the given equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ makes this possible. By rearranging the terms and applying trigonometric identities, the equation simplifies to $\frac{dy}{dx} - y \cos 2x = \cos 2x$, since $1 - 2 \sin^{2} x = \cos 2x$.

The integrating factor is calculated as $e^{\int -\cos 2x \, dx} = e^{-\frac{1}{2} \sin 2x}$. Multiplying the entire equation by this factor allows the left side to be written as the derivative of the product $y e^{-\frac{1}{2} \sin 2x}$. Integrating both sides gives $y e^{-\frac{1}{2} \sin 2x} = \int \cos 2x e^{-\frac{1}{2} \sin 2x} \, dx + c$. Evaluating the integral on the right using substitution results in $-e^{-\frac{1}{2} \sin 2x} + c$.

Substituting the condition $y\left(\frac{\pi}{4}\right)=0$ into this general solution leads to $0 = -e^{-\frac{1}{2} \sin(\frac{\pi}{2})} + c$, which simplifies to $c = e^{-\frac{1}{2}}$. With the constant known, setting $x=0$ results in $y(0) e^0 = -e^0 + e^{-\frac{1}{2}}$, or $y(0) = -1 + e^{-\frac{1}{2}}$. Consequently, adding $1$ to $y(0)$ and squaring the result yields $e^{-1}$.

Substitution of $u = y - x$ transforms the differential equation $\frac{dy}{dx}=1+x e^{y-x}$ into $\frac{du}{dx} + 1 = 1 + x e^u$, which simplifies to the separable form $\frac{du}{dx} = x e^u$. Integrating $e^{-u} du = x dx$ on both sides results in $-e^{-u} = \frac{x^2}{2} + C$. Replacing $u$ with $y-x$ yields $-e^{x-y} = \frac{x^2}{2} + C$, and applying the initial condition $y(0)=0$ identifies the constant as $C = -1$. Rearranging $-e^{x-y} = \frac{x^2}{2} - 1$ provides the explicit solution $y = x - \ln\left(\frac{2-x^2}{2}\right)$.

The derivative $\frac{dy}{dx} = 1 + x e^{y-x}$ simplifies by using $e^{y-x} = \frac{2}{2-x^2}$ to yield $\frac{dy}{dx} = \frac{2+2x-x^2}{2-x^2}$. Setting this expression equal to zero identifies the critical points where $2+2x-x^2 = 0$, or $x^{2-}2x-2 = 0$, which gives $x = 1 \pm \sqrt{3}$. Only $x = 1-\sqrt{3}$ falls within the interval $(-\sqrt{2}, \sqrt{2})$. Substituting this value into the expression for $y$ results in $(1-\sqrt{3}) - \ln\left(\frac{2-(1-\sqrt{3})^2}{2}\right)$, which simplifies to $(1-\sqrt{3})-\ln(\sqrt{3}-1)$.

The differential equation $x \tan\left(\frac{y}{x}\right) dy = \left(y \tan\left(\frac{y}{x}\right)-x\right) dx$ represents a homogeneous system where the substitution $y=vx$ effectively relates the variables $x$ and $y$. Rearranging the original expression to isolate the derivative results in $\frac{dy}{dx} = \frac{y}{x} - \cot\left(\frac{y}{x}\right)$, which simplifies the relationship significantly after applying the substitution $y=vx$ and its associated derivative $\frac{dy}{dx} = v+x\frac{dv}{dx}$.

Substituting this expression into the equation yields $v+x\frac{dv}{dx} = v-\cot(v)$, which reduces to the separable differential equation form $\tan(v) dv = -\frac{1}{x} dx$. Integrating both sides leads to the general solution $\ln|\sec(v)| = -\ln|x| + c$. By applying the initial condition $y\left(\frac{1}{2}\right)=\frac{\pi}{6}$, we determine that the constant $c=0$, which simplifies the particular solution to $y = x \cos^{-1}(x)$.

Evaluating the area of the region bounded by $x=0$, $x=\frac{1}{\sqrt{2}}$, and the curve $y=x \cos^{-1}(x)$ requires computing the definite integral $\int_{0}^{1/\sqrt{2}} x \cos^{-1}(x) \, dx$. Using integration by parts with $u = \cos^{-1}(x)$ and $dv = x \, dx$ provides the anti-derivative $\frac{x^2}{2} \cos^{-1}(x) + \frac{1}{4} \sin^{-1}(x) - \frac{1}{4} x \sqrt{1-x^2}$. Evaluating this expression from $0$ to $\frac{1}{\sqrt{2}}$ yields $\frac{1}{4}\left(\frac{\pi}{4}\right) + \frac{1}{4}\left(\frac{\pi}{4}\right) - \frac{1}{4}\left(\frac{1}{\sqrt{2}}\right)\left(\frac{1}{\sqrt{2}}\right)$, which simplifies to $\frac{1}{8}(\pi-1)$.

This differential equation, $(2 x y^{2} - y) dx + x dy = 0$, can be reorganized as $\frac{dy}{dx} - \frac{y}{x} = -2y^2$, classifying it as a Bernoulli equation. Dividing the entire equation by $y^2$ transforms it into $y^{-2}\frac{dy}{dx} - \frac{1}{x}y^{-1} = -2$. Introducing the substitution $v = y^{-1}$ leads to the derivative $\frac{dv}{dx} = -y^{-2}\frac{dy}{dx}$, which simplifies the expression into the linear first-order differential equation $\frac{dv}{dx} + \frac{v}{x} = 2$. Determining the integrating factor $e^{\int \frac{1}{x} dx} = x$ allows for the solution $vx = \int 2x dx$, which integrates to $vx = x^2 + C$. Reverting to the original variable, we obtain the general curve equation $\frac{x}{y} = x^2 + C$.

The curve passes through the intersection of the lines $2x - 3y = 1$ and $3x + 2y = 8$. Solving this linear system by elimination, multiplying the first equation by 2 and the second by 3, yields $4x - 6y = 2$ and $9x + 6y = 24$. Summing these equations results in $13x = 26$, so $x = 2$, which leads to $y = 1$. Substituting the point $(2, 1)$ into the curve equation gives $\frac{2}{1} = 2^2 + C$, simplifying to $2 = 4 + C$, which defines the constant as $C = -2$. With the specific relationship $\frac{x}{y} = x^2 - 2$, evaluating at $x = 1$ results in $\frac{1}{y} = 1 - 2 = -1$. Consequently, $y = -1$, and the absolute value $|y(1)|$ equals 1.

Rearranging the differential equation $\cos x(3 \sin x+\cos x+3) dy = [1+y \sin x(3 \sin x+\cos x+3)] dx$ by dividing both sides by $\cos x(3 \sin x+\cos x+3)dx$ isolates the derivative to yield the linear form $\frac{dy}{dx} - y \tan x = \frac{1}{\cos x(3 \sin x+\cos x+3)}$. The integrating factor for this linear equation is $e^{\int -\tan x dx} = \cos x$, and multiplying the entire equation by this factor enables the simplification $\frac{d}{dx}(y \cos x) = \frac{1}{3 \sin x+\cos x+3}$. Applying the Weierstrass substitution $\tan \frac{x}{2} = z$, with the differentials $dx = \frac{2 dz}{1+z^2}$, $\sin x = \frac{2z}{1+z^2}$, and $\cos x = \frac{1-z^2}{1+z^2}$, the integral on the right side becomes $\int \frac{dz}{z^{2+}3z+2}$, which evaluates to $\log_e \left| \frac{z+1}{z+2} \right| + C$. Given the initial condition $y(0)=0$, the constant $C$ is found to be $\log_e 2$, resulting in the solution $y \cos x = \log_e \left| \frac{\tan(x/2)+1}{\tan(x/2)+2} \right| + \log_e 2$. Evaluating this expression at $x = \frac{\pi}{3}$, where $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\cos \frac{\pi}{3} = \frac{1}{2}$, leads to $\frac{1}{2}y = \log_e \left( \frac{1/\sqrt{3}+1}{1/\sqrt{3}+2} \right) + \log_e 2$, which simplifies to $y = 2 \log_e \left( \frac{2\sqrt{3}+10}{11} \right)$.

Recognizing that the structure of the differential equation $\frac{dy}{dx}=2(y+2\sin x-5)x-2\cos x$ hides a simpler relationship allows for a swift transformation through substitution. By rearranging the terms, the expression becomes $\frac{dy}{dx} + 2\cos x = 2x(y + 2\sin x - 5)$, where the left side is precisely the derivative of $y + 2\sin x - 5$. Setting $t = y + 2\sin x - 5$ converts the expression into the separable first-order equation $\frac{dt}{dx} = 2xt$.

Integrating this simplified form with respect to $x$ results in $\ln t = x^2 + C$. The initial condition $y(0) = 7$ allows us to determine the constant $C$ by evaluating $t$ at $x=0$, which gives $t = 7 + 2\sin(0) - 5 = 2$. Substituting this into the integrated equation, we find $\ln 2 = 0^2 + C$, confirming that $C = \ln 2$. Applying this constant leads to the general solution $\ln(y + 2\sin x - 5) = x^2 + \ln 2$.

At $x = \pi$, the expression becomes $\ln(y(\pi) + 2\sin \pi - 5) = \pi^2 + \ln 2$. Since $\sin \pi = 0$, this simplifies to $\ln(y(\pi) - 5) = \pi^2 + \ln 2$. Exponentiating both sides reveals $y(\pi) - 5 = 2e^{\pi^2}$, which yields the final result $y(\pi) = 2e^{\pi^2} + 5$.

The equation $x dy = (y + x^3 \cos x) dx$ contains the structure $x dy - y dx$, which is the numerator of the derivative of the quotient $\frac{y}{x}$. Rearranging the terms to isolate this structure and dividing both sides by $x^2$ yields $\frac{x dy - y dx}{x^2} = x \cos x dx$, which simplifies to $d(\frac{y}{x}) = x \cos x dx$. Integrating both sides leads to $\frac{y}{x} = \int x \cos x dx$, where applying integration by parts yields $x \sin x + \cos x + C$.

Given the condition $y(\pi) = 0$, substituting these values determines that $0 = \pi \sin \pi + \cos \pi + C$, which simplifies to $0 = 0 - 1 + C$, fixing the constant at $C=1$. Thus, the solution function is $y = x^2 \sin x + x \cos x + x$. Evaluating this expression at $x = \frac{\pi}{2}$ gives $y(\frac{\pi}{2}) = (\frac{\pi}{2})^2 (1) + \frac{\pi}{2} (0) + \frac{\pi}{2}$, confirming the final result is $\frac{\pi^2}{4} + \frac{\pi}{2}$.

The differential equation $f(x) + x f'(x) = x^2$ is a direct application of the product rule for differentiation, where the left side is the derivative of the product $x f(x)$ with respect to $x$. Integrating both sides with respect to $x$ leads to $x f(x) = \int x^2 dx = \frac{x^3}{3} + C$.

Substituting the coordinates of the point $(-2, 2)$ into this expression allows for the determination of the integration constant $C$. Setting $x = -2$ and $f(x) = 2$ provides $(-2)(2) = \frac{(-2)^3}{3} + C$, which simplifies to $-4 = -\frac{8}{3} + C$. Solving this equation results in $C = -\frac{4}{3}$. Substituting this constant back into the general solution yields $x f(x) = \frac{x^3}{3} - \frac{4}{3}$. Multiplying the entire equation by 3 gives $3 x f(x) = x^3 - 4$, which rearranges to:

$x^3 - 3 x f(x) - 4 = 0$

This differential equation is a linear form, allowing for the direct application of an integrating factor. The expression $(y+1) \tan^{2} x dx + \tan x dy + y dx = 0$ rearranges to the standard form $\frac{dy}{dx} + \left(\frac{\sec^2 x}{\tan x}\right)y = -\tan x$. Identifying $P(x)$ as $\frac{\sec^2 x}{\tan x}$ leads to an integrating factor of $e^{\int \frac{\sec^2 x}{\tan x} dx} = \tan x$.

Multiplying the differential equation by this factor results in the derivative form $(y \tan x)' = -\tan^2 x$. Integrating both sides leads to $y \tan x = \int (1 - \sec^2 x) dx = x - \tan x + C$, which rearranges to $y = x \cot x - 1 + C \cot x$. Applying the boundary condition $\lim_{x\to 0^{+}} xy(x) = 1$ requires evaluating $\lim_{x\to 0^{+}} (x^2 \cot x - x + C x \cot x) = 1$. Since $x \cot x$ approaches 1 as $x$ tends to 0, the expression simplifies to $0 - 0 + C = 1$, confirming that $C = 1$. Substituting $x = \frac{\pi}{4}$ into $y = x \cot x - 1 + \cot x$ yields $\frac{\pi}{4}(1) - 1 + 1 = \frac{\pi}{4}$.

The differential equation is solvable by applying the substitution $t=e^y$, which satisfies $\frac{dt}{dx} = e^y \frac{dy}{dx}$ and transforms the original expression into the linear form $\frac{dt}{dx} - 2t \sin x = -\sin x \cos^2 x$. The integrating factor for this equation is $e^{\int -2 \sin x dx} = e^{2 \cos x}$. Multiplying the linear equation by this factor allows the left side to be expressed as the derivative of the product $t e^{2 \cos x}$, which simplifies the right side to $-\sin x \cos^2 x e^{2 \cos x}$.

Integrating both sides using the substitution $u = \cos x$ leads to the general solution $t e^{2 \cos x} = e^{2 \cos x}(\frac{1}{2} \cos^2 x - \frac{1}{2} \cos x + \frac{1}{4}) + C$. Applying the initial condition $y(\frac{\pi}{2}) = 0$, where $t=1$ and $\cos(\frac{\pi}{2})=0$, gives $1 = \frac{1}{4} + C$, so $C = \frac{3}{4}$. Substituting $C$ back into the expression for $e^y$ results in $e^y = \frac{1}{2} \cos^2 x - \frac{1}{2} \cos x + \frac{1}{4} + \frac{3}{4} e^{-2 \cos x}$. Setting $x=0$ gives $e^{y(0)} = \frac{1}{4} + \frac{3}{4} e^{-2}$. Comparing this expression to $e^{y(0)} = \alpha + \beta e^{-2}$ identifies $\alpha = \frac{1}{4}$ and $\beta = \frac{3}{4}$. The calculation $4(\alpha + \beta) = 4(\frac{1}{4} + \frac{3}{4})$ yields 4.