Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ are solved by determining an integrating factor, which allows the left side to be expressed as the derivative of a product. Rearranging the provided equation $(x^{2+}4)^2 dy + (2x^3y+8xy-2) dx = 0$ requires dividing by $(x^{2+}4)^2 dx$ to isolate the derivative term. This yields the linear form $\frac{dy}{dx} + \frac{2x(x^{2+}4)}{(x^{2+}4)^2}y = \frac{2}{(x^{2+}4)^2}$, which simplifies to $\frac{dy}{dx} + \left( \frac{2x}{x^{2+}4} \right)y = \frac{2}{(x^{2+}4)^2}$.

The Integrating Factor is calculated as $e^{\int \frac{2x}{x^{2+}4} dx}$, which results in $x^{2+}4$. Multiplying the linear equation by this factor transforms it into the derivative of the product $y(x^{2+}4)$, specifically \frac{d}{dx} \[y(x^{2+}4)]$ = \frac{2}{x^{2+}4}$. Integrating both sides leads to$y(x^{2+}4) = 2 \int \frac{1}{x^{2+}2^2} dx$, which evaluates to$y(x^{2+}4) = \tan^{-1}\left( \frac{x}{2} \right) + C$.

Applying the initial condition $y(0)=0$ demonstrates that the constant $C$ must be zero, as $\tan^{-1}(0)$ is zero. Thus, the specific solution is $y(x^{2+}4) = \tan^{-1}\left( \frac{x}{2} \right)$. Substituting $x=2$ into this solution gives $y(4+4) = \tan^{-1}(1)$, meaning $8y = \frac{\pi}{4}$. Dividing by eight, the final value is $y(2) = \frac{\pi}{32}$.

Solving a differential equation of the form $\frac{dy}{dx} = \frac{x+y-2}{x-y}$ requires shifting the origin to the intersection point of the lines $x+y-2=0$ and $x-y=0$. By solving this system, one finds the intersection at $x=1$ and $y=1$, which prompts the substitution $x = X+1$ and $y = Y+1$ to transform the equation into the homogeneous form $\frac{dY}{dX} = \frac{X+Y}{X-Y}$. Setting $Y = vX$ leads to the derivative expression $\frac{dY}{dX} = v + X\frac{dv}{dX}$, simplifying the entire equation to $v + X\frac{dv}{dX} = \frac{1+v}{1-v}$.

Rearranging the terms results in $X\frac{dv}{dX} = \frac{1+v^2}{1-v}$, which is separable and can be expressed as $\int \frac{1-v}{1+v^2} \, dv = \int \frac{dX}{X}$. Performing the integration yields $\tan^{-1}(v) - \frac{1}{2} \ln(1+v^2) = \ln|X| + C$. Substituting $v = \frac{y-1}{x-1}$ and $X = x-1$ back into the equation produces the solution $\tan^{-1}\left(\frac{y-1}{x-1}\right) - \frac{1}{2} \ln\left(1+\left(\frac{y-1}{x-1}\right)^2\right) = \ln|x-1| + C$. Applying the initial condition $(2,1)$ shows that $C=0$, allowing for the identification of $\alpha = 1$ and $\beta = 2$. Consequently, the value of $5\beta + \alpha$ is $5(2) + 1 = 11$.

Solving a first-order linear differential equation involves transforming the equation into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$ and determining an integrating factor. Rearranging the given equation $(1+\cos^2 x) \frac{dy}{dx} - (\sin 2x) y = \sin x$ allows for division by $(1+\cos^2 x)$, resulting in $\frac{dy}{dx} - \frac{\sin 2x}{1+\cos^2 x} y = \frac{\sin x}{1+\cos^2 x}$. The integrating factor is found by evaluating $e^{\int P(x) \, dx}$, which simplifies to $e^{\int -\frac{\sin 2x}{1+\cos^2 x} \, dx} = e^{\ln(1+\cos^2 x)} = 1+\cos^2 x$.

Multiplying the differential equation by this integrating factor transforms the left-hand side into the derivative of the product $y(1+\cos^2 x)$, leading to the relation \frac{d}{dx} \[y(1+\cos^2 x)]$ = \sin x$. Integrating both sides yields$y(1+\cos^2 x) = \int \sin x , dx$, or$y(1+\cos^2 x) = -\cos x + C$. Utilizing the condition$f(0)=0$allows for the determination of the constant$C$by substituting$x=0$and$y=0$to obtain$0 = -1 + C$, so$C=1$. Evaluating the final expression$y(1+\cos^2 x) = -\cos x + 1$at$x=\frac{\pi}{2}$results in$y(1+0) = 0 + 1$, providing the value$f\left(\frac{\pi}{2}\right)=1$.

The differential equation $\sec^2 x\,dx+(e^{2y}\tan^2 x+\tan x)\,dy=0$ simplifies by recognizing that the substitution $t=\tan x$ transforms the relationship into a more manageable linear form. Rearranging the original equation to $\sec^2 x\frac{dx}{dy}+\tan^2 x\,e^{2y}+\tan x=0$ and applying the substitution $t=\tan x$, which implies $\frac{dt}{dy}=\sec^2 x\frac{dx}{dy}$, yields the Bernoulli differential equation $\frac{dt}{dy}+t=-t^2 e^{2y}$

Dividing this equation by $t^2$ allows for the substitution $u=\frac{1}{t}$, where $\frac{du}{dy}=-t^{-2}\frac{dt}{dy}$. This substitution produces the linear differential equation $\frac{du}{dy}-u=e^{2y}$ The integrating factor for this linear equation is $e^{\int -1 \, dy}=e^{-y}$. Multiplying the entire equation by this factor results \in $\frac{d}{dy}(u e^{-y})=e^{2y} \cdot e^{-y}=e^y$. Integrating both sides leads to $u e^{-y}=e^y+C$, which translates back to $\cot x \, e^{-y}=e^y+C$ since $u=\frac{1}{\tan x}=\cot x$.

Using the initial condition $y\left(\frac{\pi}{4}\right)=0$, we substitute $x=\frac{\pi}{4}$ and $y=0$ into the equation to find $\cot\left(\frac{\pi}{4}\right) e^0=e^{0+}C$. Because $\cot\left(\frac{\pi}{4}\right)=1$, this simplifies to $1=1+C$, confirming that $C=0$. Given the condition $y\left(\frac{\pi}{6}\right)=\alpha$, we substitute $x=\frac{\pi}{6}$ and $y=\alpha$ to get $\cot\left(\frac{\pi}{6}\right) e^{-\alpha}=e^{\alpha}$, or $\sqrt{3} e^{-\alpha}=e^{\alpha}$. This yields $e^{2\alpha}=\sqrt{3}$, and raising this value to the fourth power provides $e^{8\alpha}=(\sqrt{3})^4$, which equals 9.

The tangent line to the curve at any point $(x, y)$ is represented by the equation $Y - y = Y'(x)(X - x)$. The intercepts of this line with the coordinate axes are determined by setting $X=0$ to find the y-intercept $y - x Y'(x)$, and setting $Y=0$ to find the x-intercept $x - \frac{y}{Y'(x)}$. The area of the triangle formed by these intercepts and the origin is half the product of the magnitudes of these intercepts, which simplifies to $\frac{1}{2} (x - \frac{y}{Y'(x)})(y - x Y'(x))$. Equating this area to the expression $\frac{-y^2}{2 Y'(x)} + 1$ and simplifying the algebra leads to the relationship $2xy - x^2 Y'(x) = 2$.

Rearranging this relationship into the standard form of a linear differential equation yields $\frac{dy}{dx} - \frac{2}{x}y = -\frac{2}{x^2}$. Identification of the integrating factor as $e^{\int -\frac{2}{x} dx} = \frac{1}{x^2}$ allows us to express the derivative as $\frac{d}{dx} \left( \frac{y}{x^2} \right) = -\frac{2}{x^4}$. Integrating both sides results in $\frac{y}{x^2} = \frac{2}{3x^3} + c$, which simplifies to the general form $y = \frac{2}{3x} + cx^2$. Applying the initial condition $Y(1) = 1$, we find that $1 = \frac{2}{3} + c$, so $c = \frac{1}{3}$. Substituting $X=2$ into the equation $Y(X) = \frac{2}{3X} + \frac{X^2}{3}$ gives $Y(2) = \frac{1}{3} + \frac{4}{3} = \frac{5}{3}$, and calculating $12 Y(2)$ yields $20$.

The fundamental theorem of calculus allows us to simplify the given integral equation by differentiating both sides with respect to $x$, which leads to the following relation:
$\sqrt{1 - (y')^2} = y$
Squaring both sides results in $1 - (y')^2 = y^2$, which rearranges to $(y')^2 = 1 - y^2$. This is a separable differential equation, $\frac{dy}{\sqrt{1-y^2}} = dx$, which integrates to $\sin^{-1}(y) = x + c$. Applying the initial condition $y(0) = 0$ reveals that $c = 0$, establishing that $y = \sin(x)$. Calculating the second derivative yields $y'' = -\sin(x)$, so the expression $y'' + y + 1$ becomes $-\sin(x) + \sin(x) + 1$, which simplifies to 1.

This differential equation follows the linear first-order structure $\frac{dx}{dt} + P(t)x = Q(t)$, which is best handled by finding an integrating factor. By dividing the given equation $(t+1) dx = (2 x + (t+1)^4) dt$ by $(t+1)$, we isolate the derivative to obtain the standard form $\frac{dx}{dt} - \frac{2}{t+1}x = (t+1)^3$. The integrating factor is computed as $e^{\int -\frac{2}{t+1} dt}$, which simplifies to $\frac{1}{(t+1)^2}$.

Multiplying both sides of the standard form by this integrating factor effectively turns the left side into the derivative of the product $x \cdot \frac{1}{(t+1)^2}$. Integrating both sides with respect to $t$ results in $\frac{x}{(t+1)^2} = \int (t+1) dt$, which evaluates to $\frac{x}{(t+1)^2} = \frac{(t+1)^2}{2} + C$. Substituting the initial condition $x(0)=2$ leads to $\frac{2}{(0+1)^2} = \frac{(0+1)^2}{2} + C$, establishing that $C = \frac{3}{2}$. The specific solution for the function is $x = \frac{(t+1)^4}{2} + \frac{3}{2}(t+1)^2$. Substituting $t=1$ into this expression yields $8 + 6$, which is 14.

Recognizing the structure of the differential equation after cross-multiplication allows us to group terms into identifiable differentials. Rewriting the given equation as $(2+\alpha)x dx - \beta y dx + 2 dx = \beta x dy - (2\alpha+\beta) y dy + 4\alpha dy$ lets us isolate the product rule term $\beta(x dy + y dx)$ on one side of the equality. Integrating this expression yields $\beta xy - \frac{(2\alpha+\beta)y^2}{2} + 4\alpha y = \frac{(2+\alpha)x^2}{2} + 2x + C$, where the constant $C$ vanishes since the circle passes through the origin.

For this equation to represent a circle, the $xy$ term must be eliminated, which requires $\beta$ to be zero. With $\beta$ removed, requiring the coefficients of $x^2$ and $y^2$ to be identical leads to the equality $\frac{2+\alpha}{2} = \alpha$, which simplifies to $2+\alpha = 2\alpha$ and confirms $\alpha = 2$. Substituting these values into the expression results in $2x^2 + 2y^2 + 2x - 8y = 0$, simplifying further to $x^2 + y^2 + x - 4y = 0$. Completing the square for both variables provides $(x + \frac{1}{2})^2 + (y - 2)^2 = \frac{1}{4} + 4 = \frac{17}{4}$, identifying the radius as $\frac{\sqrt{17}}{2}$.

Separating variables allows us to transform the differential equation into the integrable form $\int \frac{dy}{y(y-3)} = \int \frac{dx}{x^{2-}4}$. Applying partial fraction decomposition transforms this into $\frac{1}{3} \int (\frac{1}{y-3} - \frac{1}{y}) dy = \frac{1}{4} \int (\frac{1}{x-2} - \frac{1}{x+2}) dx$, which integrates to the general solution $\frac{1}{3} \ln \left| \frac{y-3}{y} \right| = \frac{1}{4} \ln \left| \frac{x-2}{x+2} \right| + C$.

Substituting the initial condition $x=4$ and $y=3/2$ into the expression gives $0 = \frac{1}{4} \ln (1/3) + C$, which confirms that $C = \frac{1}{4} \ln 3$. Evaluating the function at $x=10$ leads to $\frac{1}{3} \ln \left| \frac{y-3}{y} \right| = \frac{1}{4} \ln (2/3) + \frac{1}{4} \ln 3$, which simplifies to $\frac{1}{3} \ln \left| \frac{y-3}{y} \right| = \frac{1}{4} \ln 2$. Because the slope of the curve is never zero and the curve passes through $y=3/2$, we are restricted to the interval $y \in (0,3)$, meaning $\left| \frac{y-3}{y} \right| = \frac{3-y}{y}$. Solving $\frac{3-y}{y} = 8^{1/4}$ for $y$ thus yields $y = \frac{3}{1+8^{1/4}}$.

The substitution $t=x+y$ simplifies the differential equation $\frac{dy}{dx}=2x(x+y)^{3-}x(x+y)-1$ by utilizing the relation $\frac{dt}{dx} = 1 + \frac{dy}{dx}$. Substituting this into the original expression yields $\frac{dt}{dx} - 1 = 2xt^3 - xt - 1$, which reduces to the separable form $\frac{dt}{dx} = xt(2t^{2-}1)$. Separating the variables results in $\frac{dt}{t(2t^{2-}1)} = x dx$.

To integrate this expression, we substitute $z=t^2$, which implies $2t dt = dz$. Multiplying the numerator and denominator by $t$ allows us to rewrite the integral as $\int \frac{dz}{2z(2z-1)} = \int x dx$. Applying partial fraction decomposition, the left side becomes $\frac{1}{2} \int \left( \frac{1}{2z-1} - \frac{1}{2z} \right) dz$, which evaluates to $\frac{1}{2} \ln \left| \frac{2z-1}{z} \right|$. Equating this to the integral of $x dx$, we obtain $\frac{1}{2} \ln \left| \frac{2z-1}{z} \right| = \frac{x^2}{2} + C$, or more simply, $\ln \left| \frac{2z-1}{z} \right| = x^2 + K$.

Given the initial condition $y(0)=1$, we have $t(0)=0+1=1$, and consequently $z(0)=1^2=1$. Substituting these values into the integrated equation gives $\ln|1| = 0^2 + K$, confirming that the constant $K=0$. This leads to the relationship $\frac{2z-1}{z} = e^{x^2}$. Solving for $z$ at the point where $x^2$ aligns with the conditions provided in the options, we find the final value to be $z = \frac{1}{2-\sqrt{e}}$.

This differential equation relies on recognizing the term $\log_e x - \log_e y$ as $\log_e(\frac{x}{y})$, which signals that substituting the ratio $\frac{x}{y} = t$ will simplify the entire equation. Using $x = ty$ and the corresponding derivative $\frac{dx}{dy} = t + y \frac{dt}{dy}$, the expression becomes $t + y \frac{dt}{dy} = t(\log_e t + 1)$, which simplifies directly to $y \frac{dt}{dy} = t \log_e t$. Separating the variables leads to the integral form:

$\int \frac{dt}{t \log_e t} = \int \frac{dy}{y}$

Integrating the left side yields $\log_e|\log_e t| = \log_e y + C$, where the absolute value ensures the logarithm is well-defined for all arguments. Substituting $\frac{x}{y}$ back for $t$ gives $\log_e|\log_e(\frac{x}{y})| = \log_e y + C$. Given that the curve passes through the point $(e, 1)$, substituting these values results in $\log_e|\log_e(\frac{e}{1})| = \log_e(1) + C$, which simplifies to $\log_e|1| = 0 + C$, confirming that the constant $C$ is zero. Exponentiating the resulting equation $\log_e|\log_e(\frac{x}{y})| = \log_e y$ leads to the final solution curve:

$\left| \log_e \frac{x}{y} \right| = y$

The problem presents a first-order linear differential equation, which can be rearranged as $f'(x) - \alpha f(x) = 3$. Multiplying both sides by the integrating factor $e^{-\alpha x}$ allows the equation to be expressed as the derivative of a product:
$\frac{d}{dx} \$[f(x) e^{-\alpha x}]\$ = 3 e^{-\alpha x}$
Integrating both sides yields $f(x) e^{-\alpha x} = -\frac{3}{\alpha} e^{-\alpha x} + C$, which simplifies to the general solution $f(x) = -\frac{3}{\alpha} + C e^{\alpha x}$.

Using the initial condition $f(0) = 2$, we substitute $x = 0$ into the general equation to find $2 = -\frac{3}{\alpha} + C$, which determines that $C = 2 + \frac{3}{\alpha}$. Substituting this value back into the general solution gives $f(x) = -\frac{3}{\alpha} + (2 + \frac{3}{\alpha}) e^{\alpha x}$. For the function to satisfy the limit condition $\lim_{x\to -\infty } f(x) = 1$, the constant term $-\frac{3}{\alpha}$ must equal $1$, which implies $\alpha = -3$. Replacing $\alpha$ with $-3$ results in $f(x) = 1 + (2 - 1) e^{-3x}$, simplifying to $f(x) = 1 + e^{-3x}$. Evaluating the function at $f(-\ln 2)$ leads to $1 + e^{-3(-\ln 2)}$, which equals $1 + e^{3 \ln 2}$, or $1 + 8 = 9$.

The differential equations take the standard form $\frac{du}{dt} = -ku$, which have the general solution $u(t) = u(0)e^{-kt}$. Applying the initial conditions $x(0) = 2$ and $y(0) = 1$, the specific solutions become $x(t) = 2e^{-at}$ and $y(t) = e^{-bt}$.

Given the constraint $3y(1) = 2x(1)$, substituting these functions at $t=1$ gives $3e^{-b} = 4e^{-a}$, which simplifies to $e^{a-b} = \frac{4}{3}$. To find the time $t$ where $x(t) = y(t)$, set $2e^{-at} = e^{-bt}$, which simplifies to $2 = e^{(a-b)t}$. Taking the natural logarithm of both sides leads to $\ln 2 = t(a-b)$. Substituting $a-b = \ln\left(\frac{4}{3}\right)$ and solving for $t$ yields the final result:
$t = \log_{\frac{4}{3}} 2$

Any circle passing through the origin with its center on the line $y=x$ has an equation of the form $x^2 + y^2 + gx + gy = 0$, where $g$ represents an arbitrary parameter related to the coordinates of the center at $(-g/2, -g/2)$. Implicitly differentiating this expression with respect to $x$ yields $2x + 2yy' + g + gy' = 0$, which allows for the isolation of the parameter $g$ as $g = - \left( \frac{2x+2yy'}{1+y'} \right)$. Substituting this expression back into the original circle equation results in $x^2 + y^2 - \left( \frac{2x+2yy'}{1+y'} \right) (x+y) = 0$. Multiplying the entire equation by $(1+y')$ removes the denominator and produces $(x^{2+}y^2)(1+y') - 2(x+yy')(x+y) = 0$. Expanding these products leads to $x^2 + y^2 + (x^{2+}y^2)y' - 2(x^2 + xy + xyy' + y^2y') = 0$. Gathering the terms containing $y'$ on one side and the remaining terms on the other simplifies to $(x^{2+}y^{2-}2xy-2y^2)y' = 2x^{2+}2xy-x^{2-}y^2$, which further reduces to $(x^{2-}y^{2-}2xy)y' = x^{2-}y^{2+}2xy$. Expressing $y'$ as $\frac{dy}{dx}$ reveals the final differential relationship $(x^{2-}y^{2+}2xy) dx = (x^{2-}y^{2-}2xy) dy$.

This problem requires solving a first-order linear differential equation of the form
$\frac{dy}{dx} + P(x)y = Q(x)$
which can be solved systematically by finding an integrating factor defined as $IF = e^{\int P(x) \, dx}$. Multiplying the entire equation by this factor transforms the left side into the derivative of the product $y \cdot IF$, allowing the solution to be found through direct integration.

First, simplify the given differential equation by rewriting the denominator $\sin x(\sec x - \sin x \tan x)$. Since $\sec x = \frac{1}{\cos x}$ and $\tan x = \frac{\sin x}{\cos x}$, the denominator becomes $\sin x\left(\frac{1-\sin^2 x}{\cos x}\right) = \sin x \cos x$. The original equation can then be written as
$\frac{dy}{dx} = \frac{\tan x + y}{\sin x \cos x}$
which expands to $\frac{dy}{dx} = \sec^2 x + 2y \csc 2x$.

Rearranging gives the standard linear form
$\frac{dy}{dx} - (2 \csc 2x)y = \sec^2 x$
The integrating factor is $IF = e^{\int -2 \csc 2x \, dx} = e^{-\ln(\tan x)} = \cot x$. Multiplying the differential equation by this factor yields $\frac{d}{dx}(y \cot x) = \sec^2 x \cot x$. Integrating both sides results in
$y \cot x = \int \frac{1}{\sin x \cos x} \, dx = \ln(\tan x) + c$

Applying the condition $y(\frac{\pi}{4}) = 2$ where $\tan(\frac{\pi}{4}) = 1$, we find $2 = \ln(1) + c$, so $c = 2$. The specific solution is
$y = \tan x (\ln(\tan x) + 2)$
At $x = \frac{\pi}{3}$, where $\tan x = \sqrt{3}$, substituting the values gives $y(\frac{\pi}{3}) = \sqrt{3} (\ln \sqrt{3} + 2)$, which is equivalent to $\sqrt{3} (2 + \log_e \sqrt{3})$.