Differential Equations – JEE Main MathematicsPractice Questions & PYQs

This equation is handled by treating the expression as a linear differential equation where $x$ is the dependent variable and $y$ is the independent variable, specifically by rewriting it as $\frac{dx}{dy} + \frac{x}{y^2} = \frac{1}{y^3}$. Finding the integrating factor involves calculating $e^{\int \frac{1}{y^2} dy}$, which results in $e^{-1/y}$. Multiplying the differential equation by this factor allows the left side to be expressed as the derivative of the product $x e^{-1/y}$, leading to the integral equation $x e^{-1/y} = \int \frac{1}{y^3} e^{-1/y} dy + C$.

Integrating the right side utilizes the substitution $u = -1/y$, which transforms the integral into the form $\int -u e^u du$, evaluating to $(1 - u)e^u$. Replacing $u$ with its definition in terms of $y$ yields the general solution $x = \frac{1}{y} + 1 + C e^{1/y}$. Inserting the initial conditions $x=1$ and $y=1$ into this expression results in $1 = 1 + 1 + Ce$, confirming that $C = -\frac{1}{e}$. Substituting $y=2$ into the final equation gives $x = \frac{1}{2} + 1 - \frac{1}{e} e^{1/2}$, which simplifies to $\frac{3}{2} - \frac{1}{\sqrt{e}}$.

This equation is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which relies on an integrating factor to isolate the dependent variable. First, divide the entire equation by $x$ to rewrite it as $\frac{dy}{dx} + \frac{2}{x}y = x$. The integrating factor is obtained by evaluating $e^{\int \frac{2}{x} dx}$, which yields $e^{2 \ln x}$ or simply $x^2$. Multiplying the standard form by this factor changes the left side into the derivative of the product $y \cdot x^2$, so the equation becomes $\frac{d}{dx}(y \cdot x^2) = x^3$. Integrating both sides results in $y \cdot x^2 = \frac{x^4}{4} + C$, which rearranges to $y = \frac{x^2}{4} + \frac{C}{x^2}$ after dividing through by $x^2$. Applying the initial condition $y(1) = 1$ reveals that $1 = \frac{1}{4} + C$, setting the constant $C$ to $\frac{3}{4}$. Consequently, the final result is $y = \frac{x^2}{4} + \frac{3}{4x^2}$.

Recognizing the underlying structure of the differential equation $(x^{2}+1)^{2} \frac{dy}{dx} + 2x (x^{2}+1) y = 1$ is the key to solving it efficiently. By factoring out $(x^{2+}1)$, the equation simplifies to (x^{2+}1) \[ (x^{2+}1) \frac{dy}{dx} + 2xy ]$ = 1$, which represents$(x^{2+}1) \frac{d}{dx} $[y(x^{2+}1)]$ = 1$. This reveals that the derivative of the product$y(x^{2+}1)$is equal to$\frac{1}{x^{2+}1}$. Integrating both sides with respect to$x$yields$y(x^{2+}1) = \tan^{-1}(x) + c$. Using the initial condition$y(0)=0$, substituting$x=0$shows that$0(1) = \tan^{-1}(0) + c$, which results in$c=0$. Evaluating the expression at$x=1$provides$y(1)(1^{2+}1) = \tan^{-1}(1)$, which simplifies to$2y(1) = \frac{\pi}{4}$, or$y(1) = \frac{\pi}{8}$. Substituting this result into the provided relationship$\sqrt{a} y(1) = \frac{\pi}{32}$gives$\sqrt{a} \left( \frac{\pi}{8} \right) = \frac{\pi}{32}$, which leads to$\sqrt{a} = \frac{1}{4}$and consequently$a = \frac{1}{16}$.

Solving a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ depends on identifying an integrating factor defined by $I.F. = e^{\int P(x) dx}$. For the given equation $\frac{dy}{dx} + (\frac{2x+1}{x}) y = e^{-2x}$, the coefficient $P(x)$ simplifies to $2 + \frac{1}{x}$. The integrating factor is calculated as $e^{\int (2 + 1/x) dx}$, which results in $e^{2x + \ln x}$, simplifying to $x e^{2x}$. Multiplying the entire original differential equation by this integrating factor allows the left side to be expressed as the derivative of the product $y(x e^{2x})$, resulting in $\frac{d}{dx} (y x e^{2x}) = e^{-2x} \cdot x e^{2x} = x$.

Integrating both sides of the equation $\frac{d}{dx} (y x e^{2x}) = x$ leads to $y x e^{2x} = \int x dx + C$, which simplifies to $y x e^{2x} = \frac{x^2}{2} + C$. Applying the condition $y(1) = \frac{1}{2}e^{-2}$ allows for the determination of the constant $C$ by substituting $x=1$ and $y=\frac{1}{2}e^{-2}$, yielding $\frac{1}{2}e^{-2} \cdot 1 \cdot e^2 = \frac{1^2}{2} + C$, or $\frac{1}{2} = \frac{1}{2} + C$, which confirms $C = 0$. The resulting explicit solution for the function is $y = \frac{x^2}{2 x e^{2x}} = \frac{1}{2} x e^{-2x}$.

To determine where the function decreases, we analyze its derivative, $\frac{dy}{dx} = \frac{1}{2} [e^{-2x} + x(-2e^{-2x})]$, which simplifies to $\frac{1}{2} e^{-2x} (1 - 2x)$. Because $e^{-2x}$ is always positive, the sign of the derivative is determined entirely by the factor $(1 - 2x)$. The function is decreasing when this derivative is negative, which occurs when $1 - 2x < 0$, or $x > \frac{1}{2}$. Consequently, the function is strictly decreasing throughout the interval $(\frac{1}{2}, 1)$.

Solving this differential equation relies on identifying that the left-hand side is the result of applying the product rule for differentiation. The given equation $\sin x\frac{dy}{dx}+y\cos x=4x$ represents the derivative of the product $y \sin x$ with respect to $x$, which means the expression simplifies to $\frac{d}{dx}(y \sin x) = 4x$.

Integrating both sides with respect to $x$ yields $y \sin x = 2x^2 + C$. Applying the initial condition $y\left(\frac{\pi}{2}\right) = 0$ determines the constant $C$ by substituting $x = \frac{\pi}{2}$, which results in $0 = 2\left(\frac{\pi}{2}\right)^2 + C$, establishing that $C = -\frac{\pi^2}{2}$.

With the constant identified, the specific solution becomes $y \sin x = 2x^2 - \frac{\pi^2}{2}$. Substituting $x = \frac{\pi}{6}$ allows for the evaluation of $y$: $y \sin\left(\frac{\pi}{6}\right) = 2\left(\frac{\pi}{6}\right)^2 - \frac{\pi^2}{2}$. Given that $\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$, the equation becomes $\frac{y}{2} = \frac{\pi^2}{18} - \frac{\pi^2}{2}$, which simplifies to $\frac{y}{2} = \frac{\pi^2 - 9\pi^2}{18} = -\frac{8\pi^2}{18} = -\frac{4\pi^2}{9}$. Multiplying by two confirms that $y = -\frac{8\pi^2}{9}$.

This is a homogeneous differential equation, which is solved by substituting $y = vx$ and $dy = v dx + x dv$. Substituting these into the equation $(x^2 - y^2) dx + 2xy dy = 0$ yields $(x^2 - v^2 x^2) dx + 2x(vx)(v dx + x dv) = 0$, which simplifies to $(1 - v^2) dx + 2v^2 dx + 2vx dv = 0$. Grouping the terms leads to $(1 + v^2) dx + 2vx dv = 0$, which separates into $\frac{dx}{x} + \frac{2v}{1 + v^2} dv = 0$. Integrating both sides gives $\ln |x| + \ln(1 + v^2) = C$, which simplifies to $x(1 + \frac{y^2}{x^2}) = C'$, where $C'$ is a constant. Multiplying by $x$ results in $x^2 + y^2 = C'x$. Substituting the point $(1,1)$ provides $1^2 + 1^2 = C'(1)$, confirming $C' = 2$. The equation $x^2 + y^2 = 2x$ can be rearranged as $(x - 1)^2 + y^2 = 1$, which represents a circle with radius one.

The equation for a family of ellipses centered at the origin is given by $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Since the ellipse passes through the point $(0, 3)$, substituting these coordinates into the general equation gives $\frac{0^2}{a^2} + \frac{3^2}{b^2} = 1$, which simplifies to $b^2 = 9$. Thus, the family of ellipses is represented by $\frac{x^2}{a^2} + \frac{y^2}{9} = 1$.

To find the corresponding differential equation, differentiate this expression with respect to $x$:
$\frac{2x}{a^2} + \frac{2y y'}{9} = 0$
Dividing by 2 and solving for the parameter $\frac{1}{a^2}$ gives $\frac{1}{a^2} = -\frac{y y'}{9x}$. Substituting this expression for $\frac{1}{a^2}$ back into the original ellipse equation yields:
$x^2 \left(-\frac{y y'}{9x}\right) + \frac{y^2}{9} = 1$
Multiplying the entire equation by 9 simplifies it to $-x y y' + y^2 = 9$. Rearranging the terms results in the differential equation $x y y' - y^2 + 9 = 0$. (Note: There is a sign difference in the standard result compared to the provided options; however, following the derivation based on the given parameters confirms the structure of the family.)

The integrating factor for the differential equation $\frac{dy}{dx} + 2y = f(x)$ is $e^{\int 2 dx} = e^{2x}$, which leads to the general solution $y e^{2x} = \int f(x) e^{2x} dx + C$. For $x \in [0, 1]$, $f(x)=1$, so the equation $y e^{2x} = \int e^{2x} dx + C_1$ yields $y(x) = \frac{1}{2} + C_1 e^{-2x}$. Applying the initial condition $y(0) = 0$ gives $C_1 = -1/2$, resulting in $y(x) = \frac{1}{2}(1 - e^{-2x})$ for $x \in [0, 1]$, where the value at the boundary is $y(1) = \frac{1}{2}(1 - e^{-2})$.

For $x > 1$, $f(x)=0$, so the equation $\frac{dy}{dx} + 2y = 0$ has the general solution $y(x) = C_2 e^{-2x}$. Requiring continuity at $x=1$ yields $C_2 e^{-2} = \frac{1}{2}(1 - e^{-2})$, which simplifies to $C_2 e^{-2} = \frac{e^2 - 1}{2e^2}$, meaning $C_2 = \frac{e^2 - 1}{2}$. Substituting this constant into the solution for $x > 1$ gives $y(x) = \frac{e^2 - 1}{2} e^{-2x}$, and evaluating at $x = 3/2$ results in $y(3/2) = \frac{e^2 - 1}{2} e^{-3} = \frac{e^2 - 1}{2e^3}$.

Substituting $u = y^{1/5}$ into $2x = y^{1/5} + y^{-1/5}$ results in $2x = u + \frac{1}{u}$, which simplifies to the quadratic equation $u^2 - 2xu + 1 = 0$. Solving for $u$ using the quadratic formula gives $u = x \pm \sqrt{x^{2-}1}$, which implies $y = (x \pm \sqrt{x^{2-}1})^5$. Assuming $y = (x + \sqrt{x^{2-}1})^5$, differentiating with respect to $x$ yields $\frac{dy}{dx} = 5(x + \sqrt{x^{2-}1})^4 \left(1 + \frac{x}{\sqrt{x^{2-}1}}\right) = \frac{5y}{\sqrt{x^{2-}1}}$.

Squaring this derivative produces $(x^{2-}1)\left(\frac{dy}{dx}\right)^2 = 25y^2$. Differentiating both sides with respect to $x$ gives $(x^{2-}1) \cdot 2\frac{dy}{dx}\frac{d^2y}{dx^2} + 2x\left(\frac{dy}{dx}\right)^2 = 50y\frac{dy}{dx}$. Dividing the entire equation by $2\frac{dy}{dx}$ results in $(x^{2-}1)\frac{d^2y}{dx^2} + x\frac{dy}{dx} - 25y = 0$. Comparing this result with the given differential equation $(x^{2-}1)\frac{d^2y}{dx^2} + \lambda x\frac{dy}{dx} + ky = 0$, the values $\lambda = 1$ and $k = -25$ are obtained. Thus, $\lambda + k = 1 - 25 = -24$.

This problem involves a first-order differential equation that can be solved by rearranging the terms into a standard linear form or by recognizing an exact derivative. By rewriting the equation as $y\,dx - x\,dy = 3y^2\,dy$ and dividing both sides by $y^2$, we obtain the expression $\frac{y\,dx - x\,dy}{y^2} = 3\,dy$, which is the differential form of the quotient $\frac{x}{y}$. This simplification transforms the differential equation into $d\left(\frac{x}{y}\right) = 3\,dy$.

Integrating both sides with respect to their variables leads to the general solution $\frac{x}{y} = 3y + C$. To find the specific constant of integration, we use the given condition that the curve passes through the point $(1, 1)$. Substituting these values gives $1 = 3(1) + C$, which simplifies to $C = -2$. The equation of the curve is therefore $\frac{x}{y} = 3y - 2$, or $x = 3y^2 - 2y$. Checking the provided options against this equation, we test the point $\left(-\frac{1}{3}, \frac{1}{3}\right)$ by substituting $y = \frac{1}{3}$ to get $x = 3\left(\frac{1}{3}\right)^2 - 2\left(\frac{1}{3}\right) = \frac{1}{3} - \frac{2}{3} = -\frac{1}{3}$. Since this matches the $x$-coordinate, the curve passes through this point.

Recognizing the derivative of a product is the most efficient way to approach the differential equation $(2+\sin x)\frac{dy}{dx}+(y+1)\cos x=0$. This expression is equivalent to the derivative of the product $(2+\sin x)(y+1)$, because applying the product rule produces exactly those two terms. This implies $\frac{d}{dx}[(2+\sin x)(y+1)]=0$, which simplifies to the general equation $(2+\sin x)(y+1)=C$ after integration.

Substituting the initial condition $y(0)=1$ into this expression gives $(2+\sin 0)(1+1)=C$, leading to $2 \times 2 = 4$. With the constant established as $C=4$, rearrange the expression to isolate $y$, resulting in $y=\frac{4}{2+\sin x}-1$. Evaluating the function at $x=\frac{\pi}{2}$ gives $y\left(\frac{\pi}{2}\right)=\frac{4}{2+\sin\left(\frac{\pi}{2}\right)}-1$, which simplifies to $\frac{4}{3}-1$, resulting in $\frac{1}{3}$.

The differential equation is a classic example of a Bernoulli equation, which can be linearized by substituting $y^2$ with a new variable $t$. Multiplying the entire equation by $2y$ gives $2y\frac{dy}{dx}+y^2\sec x=\tan x$, and by setting $t=y^2$, the derivative $2y\frac{dy}{dx}$ becomes $\frac{dt}{dx}$. This transformation simplifies the original equation into the linear form $\frac{dt}{dx}+t\sec x=\tan x$.

The integrating factor for this linear differential equation is calculated as $I.F.=e^{\int\sec x\,dx}=\sec x+\tan x$. Multiplying the linear equation by this factor allows the left side to be written as the derivative of the product $t(\sec x+\tan x)$, so integrating both sides leads to $t(\sec x+\tan x)=\int(\sec x+\tan x)\tan x\,dx$. Evaluating this integral results in $\int\sec x\tan x\,dx+\int\tan^2 x\,dx$, which simplifies to $\sec x+(\tan x-x)+c$.

Substituting $t=y^2$ back into the expression yields $y^2(\sec x+\tan x)=\sec x+\tan x-x+c$. Applying the initial condition $y(0)=1$ leads to $1^2(\sec 0+\tan 0)=\sec 0+\tan 0-0+c$, which simplifies to $1=1+c$, confirming that $c=0$. This gives the final result:

$y^2=1-\frac{x}{\sec x+\tan x}$

Evaluating this limit via L'Hôpital's Rule transforms the indeterminate expression into a linear differential equation. Since the limit approaches a finite value as the denominator $t-x$ vanishes, we differentiate the numerator $t^2f(x)-x^2f(t)$ and the denominator $t-x$ with respect to $t$, giving:
$\lim_{t\to x} (2t f(x) - x^2 f'(t)) = 1 \implies 2x f(x) - x^2 f'(x) = 1$
Rearranging this into the standard form for a first-order linear differential equation, $\frac{df}{dx} - \frac{2}{x} f(x) = -\frac{1}{x^2}$, we find an integrating factor of $e^{-\int \frac{2}{x} dx} = \frac{1}{x^2}$. Multiplying the differential equation by this factor results in $\frac{d}{dx} \left( \frac{f(x)}{x^2} \right) = -\frac{1}{x^4}$, and integrating both sides yields:
$\frac{f(x)}{x^2} = \frac{1}{3x^3} + C$
This simplifies to $f(x) = \frac{1}{3x} + Cx^2$. Using the initial condition $f(1) = 1$, we determine the constant $C = \frac{2}{3}$, leading to the function $f(x) = \frac{1}{3x} + \frac{2x^2}{3}$. Calculating the final value at $x = \frac{3}{2}$ gives:
$f\left(\frac{3}{2}\right) = \frac{1}{3(3/2)} + \frac{2(3/2)^2}{3} = \frac{2}{9} + \frac{3}{2} = \frac{31}{18}$

Distributing $y$ in $y(1+xy)dx=xdy$ yields $y dx + xy^2 dx = x dy$, which can be rearranged to isolate $x dy - y dx = xy^2 dx$. Dividing by $y^2$ transforms the left side into the negative differential of the quotient $\frac{x}{y}$, such that $-d\left(\frac{x}{y}\right) = x dx$. Integrating both sides leads to the general solution $-\frac{x}{y} = \frac{x^2}{2} + C$.

Applying the condition that the curve passes through $(1, -1)$ results in $-\frac{1}{-1} = \frac{1^2}{2} + C$, which simplifies to $1 = \frac{1}{2} + C$ and confirms that $C = \frac{1}{2}$. Consequently, the function is defined by $y = \frac{-2x}{x^{2+}1}$. Evaluating this expression at $x = -\frac{1}{2}$ results in $f\left(-\frac{1}{2}\right) = \frac{-2(-1/2)}{(-1/2)^2 + 1} = \frac{1}{\frac{5}{4}}$, which equals $\frac{4}{5}$.

This problem involves a first-order linear differential equation, which requires isolating the derivative to reach the standard form $\frac{dy}{dx} + P(x)y = Q(x)$. Dividing the expression $(x \ln x)\frac{dy}{dx} + y = 2x \ln x$ by $(x \ln x)$ provides $\frac{dy}{dx} + \frac{y}{x \ln x} = 2$. The integrating factor is found by evaluating $e^{\int \frac{1}{x \ln x} dx}$, which simplifies cleanly to $\ln x$.

Multiplying the standard differential equation by the integrating factor allows for direct integration, resulting in $y \ln x = \int 2 \ln x \, dx$. The integral of $2 \ln x$ is $2(x \ln x - x) + C$. Evaluating this at $x=1$, where $\ln 1 = 0$, reveals that $0 = 2(0 - 1) + C$, so $C = 2$. Substituting $x=e$ into the full solution $y \ln x = 2(x \ln x - x + 1)$ gives $y(1) = 2(e - e + 1)$, which confirms that $y(e) = 2$.

The behavior of the rabbit population is described by a separable differential equation, which allows us to find the specific growth function by isolating variables and integrating. Starting with the given relationship $\frac{dp}{dt} = \frac{1}{2} p - 200$, we first factor the right side to get $\frac{dp}{dt} = \frac{1}{2}(p - 400)$, which then rearranges into the separable form $\frac{dp}{p - 400} = \frac{1}{2} dt$. Integrating both sides introduces a constant of integration, yielding $\ln|p - 400| = \frac{t}{2} + C$, which upon exponentiation becomes $p - 400 = k e^{t/2}$ where $k$ is an arbitrary constant. By substituting the initial condition $p(0) = 100$, we solve for $k$ as $100 - 400 = k e^0$, resulting in $k = -300$. Inserting this back into the expression gives $p - 400 = -300 e^{t/2}$, which simplifies to the final solution $p(t) = 400 - 300 e^{t/2}$.

Modeling the population decline requires solving the separable differential equation $\frac{d p(t)}{d t} = 0.5 p(t) - 450$. Factoring the expression on the right simplifies the setup to $\frac{d p(t)}{d t} = 0.5(p(t) - 900)$, which separates into $\frac{d p(t)}{p(t) - 900} = 0.5 dt$. Integrating both sides yields the relation $\ln |p(t) - 900| = 0.5t + C$.

Applying the initial condition $p(0) = 850$ shows that $\ln |850 - 900| = C$, which gives $C = \ln 50$. Substituting this constant and solving for $p(t)$ leads to the population function $p(t) = 900 - 50 e^{0.5t}$. To find when the population reaches zero, set $0 = 900 - 50 e^{0.5t}$ and rearrange to $18 = e^{0.5t}$. Taking the natural logarithm of both sides results in $0.5t = \ln 18$, and doubling that value gives the final time $t = 2 \ln 18$.

The value of the equipment at any point in time is determined by integrating its rate of change, which is defined by the provided differential equation. Integrating $\frac{d V(t)}{d t} = -k(T-t)$ with respect to $t$ yields the function $V(t) = \frac{k(T-t)^2}{2} + C$. By applying the initial condition that the purchase value is $I$ at $t = 0$, we establish that $I = \frac{kT^2}{2} + C$, which allows us to identify the integration constant as $C = I - \frac{kT^2}{2}$. The scrap value corresponds to the equipment's value at the end of its useful life, so evaluating the function at $t = T$ reduces the expression to $V(T) = \frac{k(T-T)^2}{2} + C$, leaving only the constant value of $I - \frac{kT^2}{2}$.