Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Determining the function $g(x)$ begins by differentiating the integral equation $\int_{0}^{x} g(t) dt = x - \int_{0}^{x} t g(t) dt$ with respect to $x$. This operation yields the relationship $g(x) = 1 - x g(x)$, which simplifies to $g(x) = \frac{1}{1+x}$.

Substituting $g(x)$ into the differential equation $\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x)$ transforms the expression into $\frac{dy}{dx} - y \tan x = 2 \sec x$. The integrating factor for this linear differential equation is $e^{\int -\tan x \, dx} = \cos x$. Applying this factor to the equation results in $\cos x \frac{dy}{dx} - y \sin x = 2$, which is equivalent to the derivative $\frac{d}{dx}(y \cos x) = 2$.

Integration of both sides gives $y \cos x = 2x + C$. Given the initial condition $y(0) = 0$, the constant $C$ must be $0$, simplifying the relation to $y \cos x = 2x$. At the value $x = \frac{\pi}{3}$, this becomes $y (\frac{1}{2}) = \frac{2\pi}{3}$, which solves to $y = \frac{4\pi}{3}$.

The equation $\frac{dx}{dy} = \frac{1+x-y^2}{y}$ is a linear differential equation that can be rewritten in the standard form $\frac{dx}{dy} - \frac{x}{y} = \frac{1-y^2}{y}$. By identifying the integrating factor as $e^{\int -\frac{1}{y} dy} = \frac{1}{y}$, we multiply the entire expression to simplify the left side into the derivative $\frac{d}{dy}(\frac{x}{y}) = \frac{1-y^2}{y^2} = y^{-2} - 1$. Integrating both sides with respect to $y$ yields $\frac{x}{y} = -\frac{1}{y} - y + C$, which simplifies to $x = -1 - y^2 + Cy$. Applying the condition $x(1) = 1$ gives $1 = -1 - 1 + C$, indicating $C = 3$, so substituting $y=2$ into the specific solution $x = -1 - y^2 + 3y$ results in $x(2) = 1$, making $5x(2) = 5$.

The solution to a first-order linear differential equation is found by identifying the integrating factor, which is the exponential of the integral of the coefficient of $y$. In this specific case, $\frac{dy}{dx}-y=1+4\sin x$, the coefficient of $y$ is $-1$, making the integrating factor $e^{-x}$. Multiplying both sides by this factor transforms the equation into the derivative of the product $y e^{-x}$, which allows us to integrate the right side, $\int (e^{-x} + 4e^{-x} \sin x) dx$. The integral of $e^{-x}$ is $-e^{-x}$, and the term $4\int e^{-x} \sin x dx$ results in $-2e^{-x}(\sin x + \cos x)$ after integration by parts. Combining these gives $y e^{-x} = -e^{-x} - 2e^{-x}(\sin x + \cos x) + C$, which simplifies to $y = -1 - 2(\sin x + \cos x) + C e^x$.

Using the boundary condition $y(\pi) = 1$, we substitute $x = \pi$ into the general solution to find $1 = -1 - 2(\sin \pi + \cos \pi) + C e^\pi$. Because $\sin \pi = 0$ and $\cos \pi = -1$, this reduces to $1 = -1 + 2 + C e^\pi$, or $1 = 1 + C e^\pi$, confirming that $C = 0$. Consequently, the specific function is $y = -1 - 2(\sin x + \cos x)$. Calculating the value at $x = \frac{\pi}{2}$ gives $y\left(\frac{\pi}{2}\right) = -1 - 2(\sin \frac{\pi}{2} + \cos \frac{\pi}{2})$, which equals $-1 - 2(1 + 0)$, resulting in $-3$. Adding $10$ to this value produces $7$.

When dealing with a differential equation of the form $\frac{dy}{dx} = (x+y+2)^2$, introducing a change of variables simplifies the structure significantly. Defining $v = x + y + 2$ leads to $\frac{dv}{dx} = 1 + \frac{dy}{dx}$, which transforms the equation into $\frac{dv}{dx} = 1 + v^2$. Separating variables and integrating both sides results in:
$\int \frac{dv}{1 + v^2} = \int dx$
This yields $\tan^{-1}(v) = x + C$.

Using the initial condition $y(0) = -2$, we find $v = 0 + (-2) + 2 = 0$ at $x = 0$. Consequently, $\tan^{-1}(0) = 0 + C$, so $C = 0$. Reverting to the original variables, $\tan^{-1}(x + y + 2) = x$ simplifies to $y = \tan(x) - x - 2$. The derivative $\frac{dy}{dx} = \sec^2(x) - 1 = \tan^2(x)$ is non-negative for all $x$ in the interval $[0, \frac{\pi}{3}]$, which indicates that the function is monotonically increasing.

Since the function is increasing, the minimum value $\beta$ occurs at the left endpoint $x = 0$, giving $\beta = \tan(0) - 0 - 2 = -2$. The maximum value $\alpha$ occurs at the right endpoint $x = \frac{\pi}{3}$, which yields $\alpha = \tan\left(\frac{\pi}{3}\right) - \frac{\pi}{3} - 2 = \sqrt{3} - \frac{\pi}{3} - 2$. Substituting these values into the expression $3\alpha + \pi$ gives:
$3\alpha + \pi = 3\left(\sqrt{3} - \frac{\pi}{3} - 2\right) + \pi = 3\sqrt{3} - 6$
Squaring this result produces $(3\sqrt{3} - 6)^2 = 27 - 36\sqrt{3} + 36 = 63 - 36\sqrt{3}$. Adding $\beta^2 = (-2)^2 = 4$ leads to a total of $67 - 36\sqrt{3}$. Comparing this to $\gamma + \delta\sqrt{3}$ identifies $\gamma = 67$ and $\delta = -36$, so $\gamma + \delta$ equals $31$.

The derivative of the product $(x^{2-}2x)\sin x$ is exactly $2(x-1)\sin x + (x^{2-}2x)\cos x$, which matches the rearranged form of the differential equation $\sec x dy + \{2(1-x) \tan x + x(2-x)\} dx = 0$, given by:

$\frac{dy}{dx} = 2(x-1)\sin x + (x^{2-}2x)\cos x.$

Integrating both sides with respect to $x$ yields the general solution $y(x) = (x^{2-}2x)\sin x + C$. Substituting the initial condition $y(0)=2$ results in $C=2$. With the specific solution $y(x) = (x^{2-}2x)\sin x + 2$ now defined, evaluating the function at $x=2$ gives $y(2) = (4-4)\sin(2) + 2$, which equals 2.

Identifying the differential equation $\frac{dy}{dx} + \frac{2x}{(1+x^2)^2} y = x e^{\frac{1}{1+x^2}}$ as a first-order linear form, the integrating factor is found by evaluating $e^{\int \frac{2x}{(1+x^2)^2} dx}$, which results in $e^{-\frac{1}{1+x^2}}$. Multiplying the entire equation by this factor reduces the left side to the derivative of $y e^{-\frac{1}{1+x^2}}$ and simplifies the right side, leading to $y e^{-\frac{1}{1+x^2}} = \frac{x^2}{2} + C$. The condition $y(0)=0$ dictates that $C=0$, and applying this to the expression for the curve results in $f(x) = \frac{x^2}{2}$. Integrating the difference between the line $y=x+4$ and the parabola $y=\frac{x^2}{2}$ over the interval from -2 to 4 provides the total enclosed area of 18.

Recognizing that the expression $(e^y+1) \cos x dx + e^y \sin x dy = 0$ is the total differential of the product $(e^y+1) \sin x$ simplifies the integration process immediately. Integrating both sides yields the general solution:

$(e^y+1) \sin x = C$

Applying the condition that the curve passes through $(\frac{\pi}{2}, 0)$ enables the determination of the constant, as $(e^0 + 1) \sin(\frac{\pi}{2}) = (1 + 1)(1) = 2$, which confirms $C = 2$. Substituting $x = \frac{\pi}{6}$ into the equation $(e^y + 1) \sin x = 2$ and utilizing the trigonometric identity $\sin(\frac{\pi}{6}) = \frac{1}{2}$ results in the relation $(e^y + 1) \frac{1}{2} = 2$. Multiplying both sides by $2$ gives $e^y + 1 = 4$, which leads to the final value of $e^y = 3$.

This is a first-order linear differential equation that is solved by identifying the integrating factor $I.F. = e^{\int 2 \, d x} = e^{2x}$ and multiplying the entire equation by it, resulting in the derivative of the product $\frac{d}{d x}(y \cdot e^{2x}) = e^{2x} \sin(2x)$. Integrating the right side yields $y \cdot e^{2x} = \frac{e^{2x}}{8}(2 \sin(2x) - 2 \cos(2x)) + C$, and applying the condition $y(0) = \frac{3}{4}$ allows us to determine that $C = 1$. The specific solution is then $y = \frac{1}{4}(\sin(2x) - \cos(2x)) + e^{-2x}$, and evaluating this at $x = \frac{\pi}{8}$ eliminates the trigonometric component because $\sin(\frac{\pi}{4}) = \cos(\frac{\pi}{4})$, leaving $e^{-\frac{\pi}{4}}$.

Linear first-order differential equations can be systematically solved by identifying an appropriate integrating factor, which allows the expression to be written as the derivative of a product. For the given equation $f'(x) - 3f(x) = \alpha$, the integrating factor is calculated as $e^{\int -3 \, dx} = e^{-3x}$.

Multiplying the differential equation by $e^{-3x}$ transforms it into the form \frac{d}{dx} \[f(x) e^{-3x}]$ = \alpha e^{-3x}$. Integrating both sides with respect to$x$results in$f(x) e^{-3x} = -\frac{\alpha}{3} e^{-3x} + C$. Multiplying throughout by$e^{3x}$gives the general solution$f(x) = C e^{3x} - \frac{\alpha}{3}$. Because$e^{3x}$approaches$0$as$x$tends toward$-\infty $, the constant term must represent the limit, so$-\frac{\alpha}{3} = 7$.

Substituting the initial condition $f(0) = 1$ into $f(x) = C e^{3x} + 7$ leads to $1 = C + 7$, which confirms that $C = -6$. The function is therefore defined by $f(x) = 7 - 6e^{3x}$. Evaluating this function at $x = -\ln 3$ gives $f(-\ln 3) = 7 - 6e^{-3 \ln 3} = 7 - 6(3^{-3}) = 7 - \frac{6}{27}$, which simplifies to $7 - \frac{2}{9}$. Multiplying this value by $9$ yields $9 \times 7 - 9 \times \frac{2}{9} = 63 - 2 = 61$.

Extracting the function $f(x)$ from the given area expression involves applying the fundamental theorem of calculus to the integral $\int_0^a f(x) dx = e^{-a} + 4a^2 + a - 1$. Differentiating this expression with respect to the variable $a$ yields $f(a) = -e^{-a} + 8a + 1$, which confirms the form of the function is $f(x) = -e^{-x} + 8x + 1$.

With $f(x)$ defined, the general solution $y = c_1 f(x) + c_2$ can be converted into a differential equation by eliminating the arbitrary constants $c_1$ and $c_2$ through successive differentiation. The first derivative is $\frac{dy}{dx} = c_1 (e^{-x} + 8)$, and the second derivative is $\frac{d^2 y}{dx^2} = c_1 (-e^{-x})$. Isolating $c_1$ from the second derivative equation gives $c_1 = -e^x \frac{d^2 y}{dx^2}$. Substituting this result into the first derivative expression leads to $\frac{dy}{dx} = (-e^x \frac{d^2 y}{dx^2}) (e^{-x} + 8)$. Expanding the terms on the right side produces $\frac{dy}{dx} = -\frac{d^2 y}{dx^2} - 8 e^x \frac{d^2 y}{dx^2}$, which simplifies to the final differential equation $(8 e^x + 1) \frac{d^2 y}{dx^2} + \frac{dy}{dx} = 0$.

A first-order linear differential equation can be solved by arranging it into the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. By dividing the given equation by $(1-x^2)dx$, we rearrange the terms to $\frac{dy}{dx} - \frac{xy}{1-x^2} = \frac{(x^{3+}2)\sqrt{3}}{\sqrt{1-x^2}}$. The integrating factor is obtained through $e^{\int P(x) dx}$, where $P(x) = -\frac{x}{1-x^2}$, resulting in $e^{\frac{1}{2}\ln(1-x^2)} = \sqrt{1-x^2}$.

Multiplying the differential equation by this integrating factor reduces the left-hand side to the derivative of $y\sqrt{1-x^2}$ and simplifies the right-hand side to $\sqrt{3}(x^{3+}2)$. Integrating both sides with respect to $x$ yields $y\sqrt{1-x^2} = \sqrt{3}(\frac{x^4}{4} + 2x) + C$. The initial condition $y(0)=0$ dictates that $C=0$. Evaluating the expression at $x = \frac{1}{2}$ gives $y(\frac{1}{2})\sqrt{1-(\frac{1}{2})^2} = \sqrt{3}(\frac{(1/2)^4}{4} + 2(1/2))$, which simplifies to $y(\frac{1}{2})\frac{\sqrt{3}}{2} = \sqrt{3}(\frac{65}{64})$. This results in $y(\frac{1}{2}) = \frac{65}{32}$. Since $m=65$ and $n=32$ are coprime, their sum $m+n$ equals 97.

Linear differential equations following the form $\frac{dy}{dx} + P(x)y = Q(x)$ rely on finding an integrating factor, $\text{IF} = e^{\int P(x) \, dx}$, to transform the left side into the derivative of a product. Dividing the given differential equation by $(2 x \ln x)$ leads to $\frac{dy}{dx} + \frac{1}{x \ln x} y = \frac{3}{2x^2}$, identifying the integrating factor as $e^{\int \frac{1}{x \ln x} \, dx}$, which simplifies directly to $\ln x$.

Multiplying the simplified differential equation by $\ln x$ allows for the integration of both sides, resulting in $y \ln x = \int \frac{3 \ln x}{2x^2} \, dx$. Evaluating the integral on the right using integration by parts yields $y \ln x = -\frac{3 \ln x}{2x} - \frac{3}{2x} + C$. Substituting the condition $y(e^{-1}) = 0$ into this expression confirms that $C = 0$. With the specific solution $y \ln x = -\frac{3 \ln x}{2x} - \frac{3}{2x}$, we evaluate at $x = e$, where $\ln x = 1$, to obtain $-\frac{3}{2e} - \frac{3}{2e}$, which simplifies to $-\frac{3}{e}$.

Newton's law of cooling describes how the temperature of an object changes relative to the surrounding environment, modeled here by the differential equation $\frac{dT}{dt}=-K(T-80)$. By separating variables and integrating, we arrive at the general solution $\ln|T-80|=-Kt+C$. Using the starting condition $T(0)=160^{\circ}\ \mathrm{F}$, the constant of integration $C$ evaluates to $\ln(80)$, which simplifies the expression to $T(t)=80+80e^{-Kt}$.

We then utilize the condition $T(15)=120^{\circ}\ \mathrm{F}$ to determine the exponential decay, leading to $120=80+80e^{-15K}$. This simplifies to $40=80e^{-15K}$, meaning $e^{-15K}=\frac{1}{2}$. To determine $T(45)$, we evaluate $80+80e^{-45K}$, which can be written as $80+80(e^{-15K})^3$. Replacing $e^{-15K}$ with $\frac{1}{2}$ gives $80+80(\frac{1}{8})$, resulting in a final value of $90^{\circ}\ \mathrm{F}$.

Separating variables is the standard strategy for differential equations of this form, which allows us to isolate the terms into the expression $dy = \frac{2x^{2+}2x+3}{x^{4+}2x^{3+}3x^{2+}2x+2} dx$. The denominator can be simplified by factoring it into the product $(x^{2+}1)(x^{2+}2x+2)$, which allows us to decompose the fraction into the simpler sum $\frac{1}{x^{2+}2x+2} + \frac{1}{x^{2+}1}$. Integrating these terms with respect to $x$ yields the general solution $y = \tan^{-1}(x+1) + \tan^{-1} x + C$.

We find the constant $C$ by substituting the given condition $y(-1) = -\frac{\pi}{4}$ into the general solution, which leads to $-\frac{\pi}{4} = \tan^{-1}(0) + \tan^{-1}(-1) + C$. Since $\tan^{-1}(0)$ is 0 and $\tan^{-1}(-1)$ is $-\frac{\pi}{4}$, the equation simplifies to $-\frac{\pi}{4} = -\frac{\pi}{4} + C$, confirming that $C = 0$. Evaluating the function at $x = 0$ then gives $y(0) = \tan^{-1}(1) + \tan^{-1}(0)$, resulting in a final value of $\frac{\pi}{4}$.

Differential equations containing a ratio of variables like $\frac{y}{x}$ are classified as homogeneous and are best resolved by substituting $v = \frac{y}{x}$ to effectively separate the variables. Starting with the equation
$x \cos \left(\frac{y}{x}\right) \frac{dy}{dx} = y \cos \left(\frac{y}{x}\right) + x$
rearrange the terms to isolate the derivative component, resulting in
$\cos \left(\frac{y}{x}\right) \left(x \frac{dy}{dx} - y\right) = x$
Dividing the entire equation by $x^2$ is helpful because it allows the left side to be expressed as the derivative of the quotient $v = \frac{y}{x}$, leading to
$\cos \left(\frac{y}{x}\right) \left(\frac{x \frac{dy}{dx} - y}{x^2}\right) = \frac{1}{x}$
Substituting $v = \frac{y}{x}$ transforms this into $\cos(v) \frac{dv}{dx} = \frac{1}{x}$, which integrates to $\sin(v) = \log_e |x| + C$. Substituting $v$ back gives $\sin \left(\frac{y}{x}\right) = \log_e |x| + C$. Applying the condition $y(1) = \frac{\pi}{3}$ yields $\sin \left(\frac{\pi}{3}\right) = \log_e(1) + C$, which simplifies to $C = \frac{\sqrt{3}}{2}$. Comparing this constant to the form $\frac{\alpha}{2}$ provided, we find $\frac{\alpha}{2} = \frac{\sqrt{3}}{2}$, which implies $\alpha = \sqrt{3}$. Consequently, $\alpha^2 = 3$.

Recognizing the structure of $x\,dy - y\,dx$ suggests that dividing the differential equation $x\,dy - y\,dx + x y (x\,dy + y\,dx) = 0$ by the product $xy$ simplifies the expression into manageable logarithmic and linear differentials. This division results in $\frac{dy}{y} - \frac{dx}{x} + d(xy) = 0$, which integrates directly to $\ln|y| - \ln|x| + xy = C$.

Substituting the initial condition $y(1) = 2$ allows for the determination of the integration constant, where $\ln|2| - \ln|1| + (1)(2) = C$ leads to $C = \ln 2 + 2$. Inserting this constant back into the general solution yields $\ln|y| - \ln|x| + xy = \ln 2 + 2$. Rearranging the equation to isolate terms gives $\ln|x| = \ln|y| - \ln 2 + xy - 2$, which simplifies further to $\ln|x| = \ln|y/2| + xy - 2$. Exponentiating both sides results in $|x| = |y/2| e^{xy - 2}$, which is equivalent to $2|x| = |y| e^{xy - 2}$. Comparing this result to the form $\alpha|x| = |y| e^{xy - \beta}$ confirms that $\alpha = 2$ and $\beta = 2$, and their sum is $\alpha + \beta = 4$.

By grouping the terms containing $\frac{dy}{dx}$ in the differential equation $2 y \frac{dy}{dx}+3=5 \frac{dy}{dx}$, the equation can be rearranged into $(2y-5) dy = -3 dx$. Integrating both sides yields $y^2 - 5y = -3x + \lambda$. Since the curve passes through the point $(0,1)$, substituting these coordinates into the expression gives $1^2 - 5(1) = -3(0) + \lambda$, which determines that $\lambda = -4$.

The resulting equation of the conic is $y^2 - 5y = -3x - 4$. Completing the square on the left side by adding $\frac{25}{4}$ results in $(y - \frac{5}{2})^2 = -3x - 4 + \frac{25}{4}$, which simplifies to the vertex form $(y - \frac{5}{2})^2 = -3(x - \frac{3}{4})$. This form indicates that the vertex of the parabola is located at $(\frac{3}{4}, \frac{5}{2})$. Substituting these vertex coordinates into the expression $2x + 3y$ gives $2(\frac{3}{4}) + 3(\frac{5}{2}) = \frac{3}{2} + \frac{15}{2} = \frac{18}{2} = 9$, confirming that the vertex lies on the line $2x+3y=9$.

Separating the variables in the differential equation $(1+y^2) e^{\tan x} dx + \cos^2 x (1+e^{2\tan x}) dy = 0$ allows us to rewrite it as $\frac{1}{1+y^2} dy + \frac{e^{\tan x} \sec^2 x}{1+e^{2\tan x}} dx = 0$. Integrating both sides leads to the expression $\tan^{-1} y + \tan^{-1}(e^{\tan x}) = C$. Applying the initial condition $y(0)=1$ gives $\tan^{-1}(1) + \tan^{-1}(e^{\tan 0}) = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2}$, which establishes that the constant $C = \frac{\pi}{2}$. With this constant determined, the equation becomes $\tan^{-1} y + \tan^{-1}(e^{\tan x}) = \frac{\pi}{2}$. Substituting $x=\frac{\pi}{4}$ results in $\tan^{-1} y + \tan^{-1}(e) = \frac{\pi}{2}$, and since $\frac{\pi}{2} - \tan^{-1}(e)$ is equivalent to $\cot^{-1} e$, the final value is $y = \frac{1}{e}$.