Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Isolating variables in the differential equation requires gathering the $dy$ terms on one side and the $dx$ terms on the other, which rearranges the expression $x^{3} dy + xy dx = x^{2} dy + 2y dx$ into $dy(x^{3} - x^{2}) = dx(2y - xy)$. Factoring the terms allows us to express this as $dy(x^{2}(x-1)) = dx(y(2-x))$, which simplifies to the separable form $-\frac{dy}{y} = \frac{x-2}{x^{2}(x-1)} dx$.

Evaluating the integral on the right side involves decomposing the integrand using partial fractions, where $\frac{x-2}{x^{2}(x-1)} = \frac{A}{x} + \frac{B}{x^{2}} + \frac{C}{x-1}$. Equating the coefficients determines that $A=1$, $B=2$, and $C=-1$, so the integral becomes $\int (\frac{1}{x} + \frac{2}{x^{2}} - \frac{1}{x-1}) dx$, resulting in $\ln x - \frac{2}{x} - \ln(x-1) + \lambda$. Setting this equal to the integral of $-\frac{dy}{y}$ yields the general solution $-\ln y = \ln x - \frac{2}{x} - \ln(x-1) + \lambda$.

Applying the condition $y(2)=e$ gives $-1 = \ln 2 - 1 - \ln(1) + \lambda$, confirming that $\lambda = -\ln 2$. Substituting this value back and evaluating the equation at $x=4$ produces $\ln y = -\ln 4 + \frac{2}{4} + \ln 3 + \ln 2$. Simplifying the right side results in $\ln y = \ln (\frac{3}{2}) + \frac{1}{2}$, which shows that $y = \frac{3}{2} \sqrt{e}$.

Separation of variables is the primary technique for identifying the curve, as it isolates the $y$ and $x$ terms effectively. Rearranging $(1+e^{-x})(1+y^{2})\frac{dy}{dx}=y^{2}$ yields $(1+y^{-2}) dy = \frac{e^x}{1+e^x} dx$, obtained by multiplying the right-hand side by $\frac{e^x}{e^x}$. Integrating both sides gives $y - \frac{1}{y} = \ln(1+e^x) + C$, and substituting the condition $(0,1)$ determines that $C = -\ln(2)$. Combining these values leads to $y - \frac{1}{y} = \ln(1+e^x) - \ln(2)$, which simplifies to $y^2 = 1 + y \ln \left(\frac{1+e^x}{2}\right)$.

The differential equation for a family of curves is determined by eliminating the arbitrary parameter through differentiation and algebraic substitution. For the curve defined by $x^2 = 4b(y + b)$, the process involves isolating the constant $b$ and rewriting the equation to relate $x$, $y$, and the derivative $y'$.

Differentiating with respect to $x$ gives $2x = 4b y'$, which simplifies to $b = \frac{x}{2y'}$. Substituting this expression back into the initial equation yields $x^2 = 4(\frac{x}{2y'})(y + \frac{x}{2y'})$. Expanding this produces $x^2 = \frac{2xy}{y'} + \frac{x^2}{(y')^2}$. Multiplying throughout by $(y')^2$ results in $x^2(y')^2 = 2xy y' + x^2$. Dividing by $x$ provides the final form $x(y')^2 = 2yy' + x$.

This problem centers on a first-order linear differential equation, which takes the standard form $y' + Py = Q$ after dividing the given equation by $x$. With $P = -1/x$ and $Q = x(x \cos x + \sin x)$, the integrating factor is $e^{\int -1/x dx}$, which simplifies to $1/x$. Multiplying both sides of the rearranged equation by this factor transforms the left side into the derivative of $y/x$, allowing for direct integration.

Integrating both sides yields $y/x = \int (x \cos x + \sin x) dx = x \sin x + C$. Substituting the boundary condition $y(\pi) = \pi$ determines that $C = 1$, resulting in the specific solution $y = x^2 \sin x + x$.

Evaluating the function at $x = \pi/2$ gives $y(\pi/2) = (\pi/2)^2 \sin(\pi/2) + \pi/2 = \pi^2/4 + \pi/2$. Differentiating twice provides $y' = 2x \sin x + x^2 \cos x + 1$ and $y'' = 2 \sin x + 4x \cos x - x^2 \sin x$. Evaluating the second derivative at $x = \pi/2$ gives $y''(\pi/2) = 2(1) + 4(\pi/2)(0) - (\pi^2/4)(1) = 2 - \pi^2/4$. Summing the function value and the second derivative results in $2 + \pi/2$.

Recognizing that the expression $(x + 1)dy - y dx$ corresponds to the numerator of the derivative of the quotient $\frac{y}{x + 1}$ allows for a direct path to the solution. By rearranging the differential equation $(x + 1)dy = ((x + 1)^2 + y - 3)dx$ into the form $(x + 1)dy - y dx = ((x + 1)^2 - 3)dx$, we can simplify the expression by dividing both sides by $(x + 1)^2$. This transformation yields the following differential relationship:
$d\left(\frac{y}{x+1}\right) = \left(1 - \frac{3}{(x+1)^2}\right)dx$
Integrating both sides leads to the general solution:
$\frac{y}{x+1} = x + \frac{3}{x+1} + C$
Substituting the initial condition $y(2) = 0$ into this equation gives $0 = 2 + \frac{3}{3} + C$, which simplifies to $0 = 3 + C$, confirming that $C = -3$. To find $y(3)$, we substitute $x = 3$ and $C = -3$ back into the general solution:
$\frac{y}{4} = 3 + \frac{3}{4} - 3$
This simplifies to $\frac{y}{4} = 0.75$, and multiplying by 4 results in $y(3) = 3.00$.

Separating variables serves as the primary technique for solving this differential equation, where terms associated with $x$ and $y$ are isolated to facilitate independent integration. Factoring the radical expression allows the equation to be rearranged into the form $\frac{y}{\sqrt{1+y^{2}}}dy = -\frac{\sqrt{1+x^{2}}}{x}dx$. The integral of the left side is $\sqrt{1+y^{2}}$. On the right side, the substitution $1+x^{2}=u^{2}$ (implying $xdx=udu$) transforms the integral into $-\int \frac{u^{2}}{u^{2}-1} du$. This expression is equivalent to $-\int (1 + \frac{1}{u^{2}-1}) du$, which integrates to $-(u + \frac{1}{2}\log_{e}|\frac{u-1}{u+1}|) + C$. Substituting $u = \sqrt{1+x^{2}}$ back into the result gives $\sqrt{1+y^{2}} = -\sqrt{1+x^{2}} - \frac{1}{2}\log_{e}|\frac{\sqrt{1+x^{2}}-1}{\sqrt{1+x^{2}}+1}| + C$. By inverting the logarithm argument to absorb the negative sign, the final expression becomes $\sqrt{1+y^{2}} + \sqrt{1+x^{2}} = \frac{1}{2}\log_{e}\left(\frac{\sqrt{1+x^{2}}+1}{\sqrt{1+x^{2}}-1}\right) + C$.

Solving differential equations of this type requires the method of separation of variables to create an integrable relationship between the two quantities. Once the variables are isolated into the form $\int \frac{dy}{2+y} = -\int \frac{e^{x}}{e^{x}+5} dx$, integrating both sides leads to the expression $\ln(y+2) = -\ln(e^{x}+5) + C$, which is equivalent to the algebraic product $(y+2)(e^{x}+5) = C$. Substituting the initial condition $y(0)=1$ gives $(1+2)(e^{0}+5) = 3(6) = 18$, defining the constant as $C=18$. Evaluating the final equation at $x=\ln 13$ results in $(y+2)(13+5) = 18$, which simplifies to $y+2=1$, yielding $y=-1$.

Linear first-order differential equations are resolved by identifying an integrating factor that allows the equation to be expressed as the derivative of a product. To begin, divide the equation $\cos x \frac{dy}{dx} + 2y \sin x = \sin 2x$ by $\cos x$ to isolate the derivative, which gives the standard linear form:
$\frac{dy}{dx} + 2y \tan x = 2 \sin x$
The integrating factor for this equation is $e^{\int 2 \tan x dx}$, which simplifies to $e^{2 \ln(\sec x)} = \sec^2 x$. Multiplying both sides of the standard form by $\sec^2 x$ results in the derivative of the product $y \sec^2 x$ on the left side:
$\frac{d}{dx} (y \sec^2 x) = 2 \sin x \sec^2 x$
Integrating both sides yields $y \sec^2 x = \int 2 \sec x \tan x dx$, which solves to $y \sec^2 x = 2 \sec x + C$. Utilizing the condition $y(\frac{\pi}{3}) = 0$, where $\sec(\frac{\pi}{3}) = 2$, leads to $0 = 2(2) + C$, confirming that $C = -4$. Substituting $x = \frac{\pi}{4}$ into the general solution $y \sec^2 x = 2 \sec x - 4$ and using $\sec(\frac{\pi}{4}) = \sqrt{2}$, we find $y(2) = 2\sqrt{2} - 4$, which simplifies to $y = \sqrt{2} - 2$.

The slope of the tangent at any point $(x, y)$ is defined by the differential equation $\frac{dy}{dx} = \frac{x^2 - 2y}{x}$. Rearranging this into the form $\frac{dy}{dx} + \frac{2}{x}y = x$ identifies it as a linear differential equation, where the integrating factor is $e^{\int \frac{2}{x} dx} = x^2$.

Multiplying the entire equation by this factor yields $\frac{d}{dx}(y x^2) = x^3$, and integrating both sides results in the general solution $y x^2 = \frac{x^4}{4} + C$. Substituting the given condition $(1, -2)$ into this expression shows that $-2 = \frac{1}{4} + C$, which determines the constant $C = -\frac{9}{4}$.

The specific equation of the curve is thus:
$y x^2 = \frac{x^4 - 9}{4}$
Testing the point $(\sqrt{3}, 0)$ in this equation leads to $0(\sqrt{3})^2 = \frac{(\sqrt{3})^4 - 9}{4}$, which simplifies to $0 = 0$. This confirms that the curve passes through the coordinates of the chosen option.

The differential equation follows the linear standard form $\frac{dy}{dx} + P(x)y = Q(x)$ by recognizing that $\frac{1}{\cos^2 x} = \sec^2 x$, which simplifies the expression to:

$\frac{dy}{dx} + 3\sec^2 x \cdot y = \sec^2 x$

Calculating the integrating factor via $e^{\int 3\sec^2 x \, dx}$ gives $e^{3\tan x}$, which transforms the differential equation into the product rule form $d(y \cdot e^{3\tan x}) = \sec^2 x \cdot e^{3\tan x} \, dx$. Integrating both sides results in:

$y \cdot e^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$

Dividing by the integrating factor yields the general solution $y = \frac{1}{3} + Ce^{-3\tan x}$. Applying the initial condition $y\left(\frac{\pi}{4}\right) = \frac{4}{3}$ results in:

$\frac{4}{3} = \frac{1}{3} + Ce^{-3(1)}$

This simplifies to $1 = Ce^{-3}$, which determines the constant $C = e^3$. Evaluating the function at $x = -\frac{\pi}{4}$ yields:

$y = \frac{1}{3} + e^3 \cdot e^{-3\tan(-\pi/4)} = \frac{1}{3} + e^3 \cdot e^{-3(-1)} = \frac{1}{3} + e^6$

The differential equation $(x^2 - y^2) dx + 2xy dy = 0$ is homogeneous because each term has a total degree of two, which indicates that the substitution $y = vx$ is an effective approach. Rearranging the original equation to $\frac{dy}{dx} = \frac{y^2 - x^2}{2xy}$ and applying the substitution $y = vx$ transforms the derivative into $v + x \frac{dv}{dx}$ on the left-hand side, leading to the following expression:

$v + x \frac{dv}{dx} = \frac{v^2 - 1}{2v}$

Subtracting $v$ from both sides simplifies the equation to $x \frac{dv}{dx} = -\frac{v^{2+}1}{2v}$. Separating the variables allows for integration of both sides, resulting in $\int \frac{2v}{v^{2+}1} dv = \int -\frac{1}{x} dx$. Calculating these integrals gives $\ln(v^{2+}1) = -\ln x + C$. Replacing $v$ with $\frac{y}{x}$ converts the expression back into Cartesian coordinates as $\ln(\frac{y^2}{x^2} + 1) = -\ln x + C$, which simplifies to $x^2 + y^2 = Cx$ after exponentiating and rearranging terms. Substituting the coordinates of the given point $(1,1)$ into this equation reveals that $C = 2$. The final equation of the curve is thus $x^2 + y^2 = 2x$, which can be rewritten in the standard form $(x-1)^2 + y^2 = 1$, identifying it as a circle with its center at $(1,0)$ on the x-axis.

Solving this differential equation involves simplifying the expression $(x-y)^2$ by utilizing a substitution that converts the equation into a separable form. By setting $t = x - y$, we find that the derivative with respect to $x$ follows the relation $\frac{dy}{dx} = 1 - \frac{dt}{dx}$. Substituting these into the original equation $\frac{dy}{dx} = (x-y)^2$ transforms the problem into $1 - \frac{dt}{dx} = t^2$, which can be rearranged as $\frac{dt}{dx} = 1 - t^2$.

Separating the variables results in $\int \frac{dt}{1-t^2} = \int dx$, and integrating both sides produces $\frac{1}{2} \ln \left| \frac{1+t}{1-t} \right| = x + \lambda$. Replacing $t$ with the original substitution $x - y$ gives $\frac{1}{2} \ln \left| \frac{1+x-y}{1-x+y} \right| = x + \lambda$. Applying the initial condition $y(1) = 1$ leads to $\frac{1}{2} \ln \left| \frac{1+1-1}{1-1+1} \right| = 1 + \lambda$, which simplifies to $\frac{1}{2} \ln(1) = 1 + \lambda$, confirming that $\lambda = -1$. Substituting this constant back into the equation results in $\frac{1}{2} \ln \left| \frac{1+x-y}{1-x+y} \right| = x - 1$, which multiplies to $\ln \left| \frac{1+x-y}{1-x+y} \right| = 2(x - 1)$. Using the logarithmic property $\ln(a) = -\ln(1/a)$, this is equivalent to $-\ln \left| \frac{1-x+y}{1+x-y} \right| = 2(x - 1)$.

Linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ are solved by determining an integrating factor, defined as $I.F. = e^{\int P(x) dx}$, which allows the equation to be condensed into a single derivative. Dividing the given equation $x \frac{dy}{dx} + y = x \ln x$ by $x$ transforms it into the standard form $\frac{dy}{dx} + \frac{1}{x} y = \ln x$, where $P(x) = \frac{1}{x}$ and $Q(x) = \ln x$. The integrating factor is calculated as $e^{\int \frac{1}{x} dx} = e^{\ln x} = x$. Multiplying the standard form by this integrating factor yields $x \frac{dy}{dx} + y = x \ln x$, which simplifies the left side to the derivative of the product $xy$, giving $\frac{d}{dx}(xy) = x \ln x$.

Integrating both sides with respect to $x$ results in $xy = \int x \ln x dx + C$. Applying integration by parts, where $\int x \ln x dx = \frac{x^2}{2} \ln x - \int \frac{x^2}{2} \cdot \frac{1}{x} dx$, the equation becomes $xy = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$. The condition $2y(2) = \ln 4 - 1$ means that at $x = 2$, $y = \frac{\ln 4 - 1}{2} = \ln 2 - 0.5$. Substituting these values into the general solution $2y = \frac{2^2}{2} \ln 2 - \frac{2^2}{4} + C$ leads to $2 \ln 2 - 1 = 2 \ln 2 - 1 + C$, confirming $C = 0$. With the constant found, the solution is $y = \frac{x}{2} \ln x - \frac{x}{4}$. Evaluating this at $x = e$ gives $y(e) = \frac{e}{2} \ln e - \frac{e}{4} = \frac{e}{2} - \frac{e}{4} = \frac{e}{4}$.

A linear differential equation of the form $\frac{dy}{dx} + \frac{3}{4x}y = 7$ describes the function $f(x)$, where $y = f(x)$. The integrating factor for this equation is $e^{\int \frac{3}{4x} dx} = x^{\frac{3}{4}}$, and multiplying the differential equation by this factor allows us to integrate both sides to find $f(x) \cdot x^{\frac{3}{4}} = \int 7x^{\frac{3}{4}} dx = 4x^{\frac{7}{4}} + C$. Dividing through by $x^{\frac{3}{4}}$ results in the general solution $f(x) = 4x + C \cdot x^{-\frac{3}{4}}$. Computing the limit of $x f(\frac{1}{x})$ involves substituting $\frac{1}{x}$ for $x$ in this general solution, which leads to $x \left(4(\frac{1}{x}) + C \cdot x^{\frac{3}{4}}\right) = 4 + C \cdot x^{\frac{7}{4}}$. As $x \to 0^{+}$, the term $C \cdot x^{\frac{7}{4}}$ approaches zero, which leaves a final value of 4.

The integrating factor method for solving linear first-order differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ uses $I(x) = e^{\int P(x) dx}$ to transform the equation into the derivative of a product. Rearranging the given equation $\frac{dy}{dx} = (\tan x - y) \sec^2 x$ into $\frac{dy}{dx} + (\sec^2 x)y = \tan x \sec^2 x$ identifies $P(x) = \sec^2 x$, which results in an integrating factor of $e^{\int \sec^2 x \, dx} = e^{\tan x}$.

Multiplying the rearranged equation by $e^{\tan x}$ results in $\frac{d}{dx}(y e^{\tan x}) = e^{\tan x} \tan x \sec^2 x$. Integrating both sides requires the substitution $u = \tan x$, where $du = \sec^2 x \, dx$, converting the right integral to $\int u e^u \, du = e^u(u-1)$. This yields the general solution $y e^{\tan x} = e^{\tan x}(\tan x - 1) + C$, which simplifies to $y = \tan x - 1 + C e^{-\tan x}$.

Applying the initial condition $y(0)=0$ leads to $0 = \tan(0) - 1 + C e^{-\tan(0)}$, so $0 = -1 + C$, implying $C=1$. Substituting $x = -\frac{\pi}{4}$ into $y = \tan x - 1 + e^{-\tan x}$ gives $y(-\frac{\pi}{4}) = \tan\left(-\frac{\pi}{4}\right) - 1 + e^{-\tan\left(-\frac{\pi}{4}\right)} = -1 - 1 + e^1 = e-2$.

This Bernoulli differential equation relies on substituting the dependent variable to transform it into a linear first-order equation. Starting with the given relation and rearranging terms to isolate the derivative results in $\frac{dy}{dx} = \frac{y}{x} - \frac{x^2}{y}$. Multiplying this entire equation by $2y$ creates a structure where $2y \frac{dy}{dx}$ acts as the derivative of $y^2$, leading to the form $2y \frac{dy}{dx} - \frac{2y^2}{x} = -2x^2$. By applying the substitution $z = y^2$, the equation becomes a linear differential equation in $z$: $\frac{dz}{dx} - \frac{2}{x}z = -2x^2$.

Solving this with the integrating factor $e^{\int -(2/x) \, dx} = x^{-2}$ provides the solution $\frac{z}{x^2} = \int -2 \, dx$, which simplifies to $\frac{z}{x^2} = -2x + c$. Multiplying through by $x^2$ and substituting back $y^2$ yields $y^2 = -2x^3 + cx^2$, which rearranges to $y^2 + 2x^3 - cx^2 = 0$. Given that $c$ is an arbitrary constant, this is equivalent to $y^2 + 2x^3 + cx^2 = 0$.

First-order linear differential equations, which take the form $\frac{dy}{dx} + P(x)y = Q(x)$, can be solved by calculating an integrating factor. Dividing the original expression $x \frac{dy}{dx} + 2y = x^2$ by $x$ yields $\frac{dy}{dx} + \frac{2}{x}y = x$. The integrating factor is found using $e^{\int \frac{2}{x} dx}$, which simplifies to $e^{2 \ln x}$, or $x^2$. Multiplying the standard form equation by this factor transforms the left side into the derivative of the product $yx^2$, leading to $\frac{d}{dx}(yx^2) = x^3$. Integrating both sides results in:
$yx^2 = \frac{x^4}{4} + C$

Applying the condition $y(1) = 1$ allows us to determine the constant by substituting $1$ for both $x$ and $y$, yielding $1 = \frac{1}{4} + C$, which confirms $C$ is $\frac{3}{4}$. The general solution becomes $yx^2 = \frac{x^4}{4} + \frac{3}{4}$. When evaluating this expression at $x = \frac{1}{2}$, we find $y(\frac{1}{2}) \cdot \frac{1}{4} = \frac{(1/2)^4}{4} + \frac{3}{4}$, which simplifies to $\frac{1}{64} + \frac{3}{4}$. Finding a common denominator, this equals $\frac{1+48}{64}$, or $\frac{49}{64}$. Multiplying both sides by $4$ to isolate $y(\frac{1}{2})$ results in a final value of $\frac{49}{16}$.

The functional equation $f(xy) = f(x)f(y)$ implies that $f(x)$ is a constant function. Substituting $y=0$ into the equation yields $f(0) = f(x)f(0)$, and since $f(0)$ is given as non-zero, we can divide both sides by $f(0)$ to conclude that $f(x) = 1$ for all $x \in [0, 1]$.

With this result, the differential equation $\frac{dy}{dx} = f(x)$ simplifies to $\frac{dy}{dx} = 1$. Integrating both sides with respect to $x$ produces $y = x + C$. Applying the initial condition $y(0) = 1$, we determine that $1 = 0 + C$, which gives $C = 1$, so the expression for $y$ becomes $y(x) = x + 1$.

To find the final value, we evaluate the expression at the given points. Calculating $y(\frac{1}{4}) + y(\frac{3}{4})$ involves summing $(\frac{1}{4} + 1)$ and $(\frac{3}{4} + 1)$, which results in $\frac{5}{4} + \frac{7}{4}$, simplifying to $3$.