Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Recognizing the differential equation as a first-order linear type allows for a straightforward solution using an integrating factor. Rearranging the expression $(x \cos x) dy+(x y \sin x+y \cos x-1) dx=0$ leads to the standard form
$\frac{dy}{dx} + \left( \tan x + \frac{1}{x} \right) y = \frac{\sec x}{x}$
where the integrating factor is $e^{\int (\tan x + \frac{1}{x}) dx} = x \sec x$. Multiplying the standardized equation by this factor simplifies the left side to the derivative of the product $y \cdot x \sec x$, which allows for direct integration to give $y \cdot x \sec x = \tan x + c$. Solving for $y$ yields $y = \frac{\sin x + c \cos x}{x}$, and applying the condition $\frac{\pi}{3} y \left(\frac{\pi}{3}\right) = \sqrt{3}$ determines that $c = \sqrt{3}$, so the solution is $y = \frac{\sin x + \sqrt{3} \cos x}{x}$. Considering the expression $xy = \sin x + \sqrt{3} \cos x$, differentiating with respect to $x$ provides $x y' + y = \cos x - \sqrt{3} \sin x$, and differentiating once more results in $x y'' + 2 y' = -(\sin x + \sqrt{3} \cos x)$. Evaluating this at $x = \frac{\pi}{6}$ gives $-\left( \sin \frac{\pi}{6} + \sqrt{3} \cos \frac{\pi}{6} \right) = - \left( \frac{1}{2} + \frac{3}{2} \right) = -2$, confirming the final magnitude is 2.

The differential equation $\frac{dy}{dx}+\frac{x+a}{y-2}=0$ can be rewritten as $(2-y)dy = (x+a)dx$ by separating variables. Integrating both sides yields $2y - \frac{y^2}{2} = \frac{x^2}{2} + ax + C$, which simplifies to $x^2 + 2ax + y^2 - 4y + 2C = 0$ after multiplying by two. Completing the square transforms this into the standard circle equation $(x+a)^2 + (y-2)^2 = r^2$, where the radius squared is $r^2 = a^2 + 4 - 2C$.

Applying the condition $y(1)=0$ to the equation $x^2 + 2ax + y^2 - 4y + k = 0$ gives $1+2a+k=0$, and equating the given area $4\pi$ to $\pi r^2$ establishes $r^2=4$. Substituting the constants reveals $a=-1$, resulting in the circle $(x-1)^2 + (y-2)^2 = 4$. Setting $x=0$ to find the intersection points with the $y$-axis yields $(y-2)^2 = 3$, meaning $y = 2 \pm \sqrt{3}$, identifying the points as $P(0, 2+\sqrt{3})$ and $Q(0, 2-\sqrt{3})$.

The normal to a circle at any point passes through its center $(1, 2)$, so the normals at $P$ and $Q$ have slopes of $-\sqrt{3}$ and $\sqrt{3}$, respectively. The equations of these normals are $y-2=-\sqrt{3}(x-1)$ and $y-2=\sqrt{3}(x-1)$. Finding the $x$-axis intercepts by setting $y=0$ yields $x = 1 - \frac{2}{\sqrt{3}}$ and $x = 1 + \frac{2}{\sqrt{3}}$. The distance between these two intercepts is $\frac{4}{\sqrt{3}}$, which simplifies to $\frac{4\sqrt{3}}{3}$.

The differential equation $\frac{dy}{dx} + y \tan x = x \sec x$ is a first-order linear differential equation, which can be solved by determining an integrating factor. By calculating $e^{\int \tan x \, dx}$, we find the integrating factor to be $\sec x$. Multiplying the entire equation by this factor transforms the left side into the derivative of the product $y \sec x$, resulting in the equation $\frac{d}{dx} (y \sec x) = x \sec^2 x$. Integrating both sides with respect to $x$ requires the use of integration by parts on the right side, which yields $y \sec x = x \tan x - \log_e(\sec x) + c$.

Applying the initial condition $y(0)=1$, we substitute $x=0$ and $y=1$ to determine that $c=1$, leading to the specific solution $y \sec x = x \tan x - \log_e(\sec x) + 1$. To evaluate the curve at $x = \frac{\pi}{6}$, we substitute $\frac{\pi}{6}$ into the equation to obtain $y (\frac{2}{\sqrt{3}}) = \frac{\pi}{6} (\frac{1}{\sqrt{3}}) - \log_e(\frac{2}{\sqrt{3}}) + 1$. Solving for $y$ by multiplying through by $\frac{\sqrt{3}}{2}$ gives $y = \frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e(\frac{2}{\sqrt{3}}) + \frac{\sqrt{3}}{2}$. By expressing the constant $\frac{\sqrt{3}}{2}$ as $\frac{\sqrt{3}}{2} \log_e e$, the expression can be rearranged as $\frac{\pi}{12} - \frac{\sqrt{3}}{2} (\log_e(\frac{2}{\sqrt{3}}) - \log_e e)$, which simplifies to $\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e \left( \frac{2}{e \sqrt{3}} \right)$.

To solve the differential equation $2(y+2)\ln(y+2)dx + (x+4-2\ln(y+2))dy = 0$, the most effective strategy is to treat $x$ as a function of $y$ and rearrange the equation into a linear differential form. By substituting $t = \ln(y+2)$, we find that $dt = \frac{dy}{y+2}$, which allows us to rewrite the differential equation as $2t \frac{dx}{dt} + x + 4 - 2t = 0$. Isolating $\frac{dx}{dt}$ gives us the standard linear form $\frac{dx}{dt} + \frac{1}{2t}x = 1 - \frac{2}{t}$.

With the equation in the form $\frac{dx}{dt} + P(t)x = Q(t)$, we determine the integrating factor as $e^{\int \frac{1}{2t} dt} = e^{\frac{1}{2}\ln t} = \sqrt{t}$. Multiplying the entire equation by $\sqrt{t}$ simplifies the left side into the derivative of the product $x\sqrt{t}$, meaning $\frac{d}{dt}(x\sqrt{t}) = \sqrt{t} - \frac{2}{\sqrt{t}}$. Integrating both sides with respect to $t$ yields $x\sqrt{t} = \frac{2}{3}t^{3/2} - 4t^{1/2} + C$, which simplifies to the general solution $x = \frac{2}{3}t - 4 + \frac{C}{\sqrt{t}}$.

To find the constant $C$, we use the initial condition $x(e^{4-}2)=1$, which corresponds to $t = \ln(e^{4-}2+2) = 4$. Substituting these values gives $1 = \frac{2}{3}(4) - 4 + \frac{C}{\sqrt{4}}$, simplifying to $1 = \frac{8}{3} - 4 + \frac{C}{2}$, which results in $C = \frac{14}{3}$. Finally, to evaluate $x$ at $e^{9-}2$, we set $t = \ln(e^{9-}2+2) = 9$ and compute $x = \frac{2}{3}(9) - 4 + \frac{14/3}{\sqrt{9}}$, which simplifies to $6 - 4 + \frac{14}{9}$, yielding $\frac{32}{9}$.

Dividing both sides of the equation $x \log_e x \frac{dy}{dx} + y = x^2 \log_e x$ by $x \ln x$ converts the expression into the standard linear form
$\frac{dy}{dx} + \frac{y}{x \ln x} = x$
The integrating factor is $I.F. = e^{\int \frac{1}{x \ln x} dx}$, which simplifies to $e^{\ln(\ln x)}$ and results in $\ln x$ since $x > 1$. Multiplying the differential equation by $\ln x$ allows the left side to be expressed as the derivative of the product $y \ln x$, leading to $\frac{d}{dx}(y \ln x) = x \ln x$.

Integrating both sides with respect to $x$ requires using integration by parts on the right, which gives
$y \ln x = \frac{x^2}{2} \ln x - \frac{x^2}{4} + C$
Substituting the initial condition $y(2) = 2$ leads to $2 \ln 2 = \frac{4}{2} \ln 2 - \frac{4}{4} + C$, which simplifies to $2 \ln 2 = 2 \ln 2 - 1 + C$, meaning the constant $C = 1$. The resulting expression $y = \frac{x^2}{2} - \frac{x^2}{4 \ln x} + \frac{1}{\ln x}$ evaluated at $x = e$ yields $y(e) = \frac{e^2}{2} - \frac{e^2}{4} + 1$, simplifying to $\frac{e^{2+}4}{4}$.

This linear differential equation utilizes the integrating factor method, where the integrating factor is $e^{\alpha t}$. Multiplying the equation by this factor transforms it into the derivative of a product, which integrates to the form $y e^{\alpha t} = \frac{\gamma e^{(\alpha-\beta)t}}{\alpha-\beta} + C$. Dividing this result by $e^{\alpha t}$ expresses the general solution as $y(t) = \frac{\gamma e^{-\beta t}}{\alpha-\beta} + C e^{-\alpha t}$. Since $\alpha > 0$ and $\beta > 0$, the exponential terms $e^{-\beta t}$ and $e^{-\alpha t}$ both decay to zero as $t$ approaches infinity, causing the limit of $y(t)$ to be zero.

The problem involves a first-order linear differential equation that simplifies through a strategic substitution. Recognizing that the term $(1+\ln x) \frac{dx}{dy}$ represents the derivative $\frac{d}{dy}(x \ln x)$, we set $t = x \ln x$ to transform the given expression into the standard linear form $\frac{dt}{dy} - t = e^y$.

Solving this linear equation requires an integrating factor, which is calculated as $e^{\int -1 dy} = e^{-y}$. Multiplying the equation by this factor results in the derivative of the product $t e^{-y}$ on the left side, giving $\frac{d}{dy}(t e^{-y}) = 1$. Integrating both sides with respect to $y$ yields $t e^{-y} = y + c$, or by substituting back for $t$, $x \ln x e^{-y} = y + c$.

Applying the condition $(1, 0)$ into the general solution gives $1 \cdot \ln(1) \cdot e^0 = 0 + c$, which confirms that $c = 0$. With $c = 0$, the equation simplifies to $x \ln x e^{-y} = y$. Substituting the coordinates of the point $(\alpha, 2)$ leads to $\alpha \ln \alpha e^{-2} = 2$, which rearranges to $\alpha \ln \alpha = 2 e^2$. Since $\alpha \ln \alpha$ is equivalent to $\ln(\alpha^\alpha)$, the final result is $\alpha^\alpha = e^{2 e^2}$.

Linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ allow for a solution via an integrating factor. Rearranging $x^3 dy + (xy - 1) dx = 0$ leads to $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3}$, which yields an integrating factor of $e^{\int \frac{1}{x^2} dx} = e^{-\frac{1}{x}}$. Multiplying the equation by this factor produces the relationship $\frac{d}{dx} (y e^{-\frac{1}{x}}) = \frac{1}{x^3} e^{-\frac{1}{x}}$.

Substituting $t = -\frac{1}{x}$ with $dt = \frac{1}{x^2} dx$ allows the integration of the right side as $-\int t e^t dt$, which evaluates to $-e^t(t-1) + C$, or $e^{-\frac{1}{x}}(1 + \frac{1}{x}) + C$ in terms of $x$. This gives the general solution $y = 1 + \frac{1}{x} + C e^{\frac{1}{x}}$. Applying the initial condition $y(\frac{1}{2}) = 3 - e$ results in $3 - e = 1 + 2 + C e^2$, identifying the constant as $C = -\frac{1}{e}$. Substituting $x=1$ into the specific solution $y = 1 + \frac{1}{x} - e^{\frac{1}{x}-1}$ yields $y(1) = 1 + 1 - e^0 = 1$.

Solving a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ involves identifying the integrating factor $\text{I.F.} = e^{\int P(x) \, dx}$, which allows us to condense the derivative of the product $y \cdot \text{I.F.}$ Here, $P(x) = \frac{5}{x(x^{5+}1)}$, and we calculate the integral by multiplying the numerator and denominator by $x^4$, resulting in $\int \frac{5x^4}{x^5(x^{5+}1)} \, dx$. Using the substitution $u = x^5$, the integral becomes $\int (\frac{1}{u} - \frac{1}{u+1}) \, du = \ln|\frac{x^5}{x^{5+}1}|$, so the integrating factor is $\frac{x^5}{x^{5+}1}$.

Multiplying both sides of the original equation by this integrating factor transforms the left side into the derivative $\frac{d}{dx} \left( y \cdot \frac{x^5}{x^{5+}1} \right)$, while the right side simplifies to $\int \frac{(x^{5+}1)^2}{x^7} \cdot \frac{x^5}{x^{5+}1} \, dx$, which reduces to $\int (x^3 + x^{-2}) \, dx$. Integrating these terms gives $y \cdot \frac{x^5}{x^{5+}1} = \frac{x^4}{4} - \frac{1}{x} + C$. Applying the condition $y(1) = 2$ leads to $2 \cdot \frac{1}{2} = \frac{1}{4} - 1 + C$, which simplifies to $1 = -0.75 + C$, fixing the constant at $C = \frac{7}{4}$. Substituting $x = 2$ into the expression results in $y \cdot \frac{32}{33} = \frac{2^4}{4} - \frac{1}{2} + \frac{7}{4} = 4 - 0.5 + 1.75 = \frac{21}{4}$, which yields $y = \frac{21}{4} \cdot \frac{33}{32} = \frac{693}{128}$.

Problems containing a variable limit of integration often require differentiation to uncover an underlying differential equation. Differentiating the given expression $f(x) + \int_{3}^{x} \frac{f(t)}{t} dt = \sqrt{x+1}$ with respect to $x$ yields $f'(x) + \frac{f(x)}{x} = \frac{1}{2\sqrt{x+1}}$. This is a linear differential equation solvable using an integrating factor of $e^{\int \frac{1}{x} dx} = x$.

Applying this factor, the equation becomes $x f'(x) + f(x) = \frac{x}{2\sqrt{x+1}}$, which simplifies to the derivative of a product $\frac{d}{dx} [x f(x)] = \frac{x}{2\sqrt{x+1}}$. Integrating both sides with the substitution $x+1 = t^2$ results in $x f(x) = \frac{(x+1)^{3/2}}{3} - \sqrt{x+1} + C$. Setting $x=3$ in the original equation shows $f(3) = 2$, which allows for the determination of $C$ by substituting into the expression to get $6 = \frac{8}{3} - 2 + C$, resulting in $C = \frac{16}{3}$. Computing the value at $x=8$ provides $8 f(8) = \frac{27}{3} - 3 + \frac{16}{3} = \frac{34}{3}$, which yields $17$ when multiplied by $12$.

Separable variables turn $y(x+1) dx - x^2 d y = 0$ into the integrable form $\int (\frac{1}{x} + x^{-2}) dx = \int \frac{dy}{y}$. Integrating both sides gives $\ln x - \frac{1}{x} + C = \ln y$, and applying the condition $y(1)=e$ yields $\ln 1 - 1 + C = 1$, which identifies the constant as $C = 2$. Exponentiating this solution leads to $y = e^{\ln x - \frac{1}{x} + 2}$, which simplifies further to $y = x e^{-\frac{1}{x} + 2}$. Since the exponential term $e^{-\frac{1}{x}}$ approaches $0$ as $x$ approaches $0^{+}$, the limit of the entire expression becomes $0$.

Homogeneous differential equations, where the numerator and denominator share the same degree, are efficiently solved using the substitution $y = vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Applying this substitution transforms the original equation into $v + x \frac{dv}{dx} = - \frac{1 + 3v^2}{3 + v^2}$, and simplifying the terms leads to $x \frac{dv}{dx} = - \frac{(v+1)^3}{3 + v^2}$. Separating variables results in the integral equation $\int \frac{3+v^2}{(v+1)^3} \, dv = - \int \frac{dx}{x}$, and decomposing the integrand into partial fractions yields $\int \left( \frac{1}{v+1} - \frac{2}{(v+1)^2} + \frac{4}{(v+1)^3} \right) \, dv = -\ln|x| + C$. Integrating these terms gives $\ln|v+1| + \frac{2}{v+1} - \frac{2}{(v+1)^2} = -\ln|x| + C$, which simplifies by combining logarithmic terms and substituting $v = \frac{y}{x}$ to arrive at $\ln|x+y| + \frac{2xy}{(x+y)^2} = C$. Given the boundary condition $y(1) = 0$, the integration constant $C$ is found to be zero, resulting in the final solution $\log_e |x+y| + \frac{2xy}{(x+y)^2} = 0$.

The differential equation $\frac{dy}{dx} = \frac{y}{x}(1+xy^2(1+\log_e x))$ is of the Bernoulli form, which can be rearranged as $\frac{dy}{dx} - \frac{y}{x} = y^3(1+\log_e x)$. Dividing the equation by $y^3$ and applying the substitution $t = -y^{-2}$ yields $\frac{dt}{dx} = 2y^{-3}\frac{dy}{dx}$, transforming the expression into the linear first-order form $\frac{dt}{dx} + \frac{2t}{x} = 2(1+\log_e x)$.

The integrating factor for this linear equation is $x^2$, and multiplying by this factor yields $\frac{d}{dx}(tx^2) = 2x^2(1+\log_e x)$. Integrating both sides results in $tx^2 = 2\left(\frac{x^3}{3} + \frac{x^3}{3}\log_e x - \frac{x^3}{9}\right) + C$, which simplifies to $tx^2 = \frac{4x^3}{9} + \frac{2x^3}{3}\log_e x + C$. Given the initial condition $y(1)=3$, we have $t(1)=-1/9$, which leads to $C=-5/9$. Substituting this constant and rearranging $-\frac{x^2}{y^2} = \frac{4x^3 + 6x^3\log_e x - 5}{9}$ gives $\frac{y^2}{9} = \frac{x^2}{5 - 2x^3(2 + 3\log_e x)}$, which, by applying $3\log_e x = \log_e x^3$, results in $\frac{x^2}{5 - 2x^3(2 + \log_e x^3)}$.

The given differential equation takes the form of a Bernoulli equation, which simplifies significantly upon dividing by $y^2$ and substituting $v = y^{-1}$. Differentiating $v$ with respect to $x$ yields $\frac{dv}{dx} = -y^{-2} \frac{dy}{dx}$, allowing the equation $(1+x^2) \frac{dy}{dx} = xy - y^2$ to be re-expressed as $\frac{dv}{dx} + v \frac{x}{1+x^2} = \frac{1}{1+x^2}$ after multiplying the original expression by $-y^{-2}$. The integrating factor for this linear equation is found by evaluating $e^{\int \frac{x}{1+x^2} dx}$, which results in $\sqrt{1+x^2}$.

Multiplying the linear equation by this factor leads to the derivative of the product $v\sqrt{1+x^2}$, which integrates to $v\sqrt{1+x^2} = \ln(x + \sqrt{x^{2+}1}) + C$. Substituting $v=y^{-1}$ back into the expression gives $\frac{\sqrt{1+x^2}}{y} = \ln(x + \sqrt{x^{2+}1}) + C$. Using the initial condition $y(0)=1$, we determine $1 = \ln(1) + C$, establishing that $C=1$. At the point $x=2\sqrt{2}$, where $y=\beta$, the term $\sqrt{1+x^2}$ evaluates to $\sqrt{1+8}=3$, leading to the relation $\frac{3}{\beta} = \ln(2\sqrt{2} + \sqrt{8+1}) + 1$. This simplifies to $\frac{3}{\beta} = \ln(2\sqrt{2}+3) + \ln e = \ln(e(3+2\sqrt{2}))$, which confirms that $e^{3 \beta^{-1}} = e(3+2\sqrt{2})$.

This problem is essentially a linear first-order differential equation, which takes the general form $\frac{dy}{dx} + P(x)y = Q(x)$. By working with the differential equation $(y-2 \ln x) dx + (2 x \ln x) dy = 0$, we can rearrange the terms to isolate the derivative, obtaining $\frac{dy}{dx} = \frac{2 \ln x - y}{2 x \ln x}$, which simplifies to $\frac{dy}{dx} + \frac{y}{2 x \ln x} = \frac{1}{x}$. The integrating factor for this linear equation is $e^{\int \frac{1}{2 x \ln x} dx}$, which resolves to $e^{\frac{1}{2} \ln(\ln x)} = (\ln x)^{1/2}$.

Multiplying the differential equation by this integrating factor results in $y \sqrt{\ln x} = \int \frac{\sqrt{\ln x}}{x} dx$. By performing a substitution where $u^2 = \ln x$ and $2u du = \frac{dx}{x}$, the integral evaluates to $\frac{2}{3} (\ln x)^{3/2} + C$. Using the given point $\left(e, \frac{4}{3}\right)$, where $\ln x = 1$, we can solve for the constant of integration:
$\frac{4}{3} = \frac{2}{3} + C \implies C = \frac{2}{3}$
With the general solution $y \sqrt{\ln x} = \frac{2}{3} (\ln x)^{3/2} + \frac{2}{3}$ now fully defined, we substitute the second point $\left(e^4, \alpha\right)$ to find $\alpha \sqrt{\ln(e^4)} = \frac{2}{3} (\ln(e^4))^{3/2} + \frac{2}{3}$. Since $\ln(e^4) = 4$, this simplifies to $\alpha \cdot 2 = \frac{2}{3} (8) + \frac{2}{3} = \frac{18}{3} = 6$. Consequently, $\alpha = 3$.

Because this differential equation is homogeneous, isolating the derivative gives $\frac{dy}{dx} = \frac{3y^{2-}x^2}{3xy}$, which simplifies to $\frac{dy}{dx} = \frac{y}{x} - \frac{1}{3}\frac{x}{y}$. Applying the standard substitution $y = vx$ leads to $v + x\frac{dv}{dx} = v - \frac{1}{3v}$, which simplifies further to the separable expression $3v \, dv = -\frac{1}{x} \, dx$.

Integrating both sides provides $\frac{3v^2}{2} = -\ln x + C$. Replacing $v$ with $y/x$ gives $\frac{3y^2}{2x^2} = -\ln x + C$. Given the initial condition $y(1) = 1$, the constant of integration is determined as $C = 3/2$. Solving for $y^2$ results in $y^2 = x^2 - \frac{2}{3}x^2 \ln x$. Finally, evaluating this function at $x = e$ yields $y^2(e) = e^2 - \frac{2}{3}e^2 = \frac{e^2}{3}$, and multiplying this value by 6 results in $2e^2$.