Differential Equations – JEE Main MathematicsPractice Questions & PYQs

The order of a differential equation is the highest derivative present after eliminating all arbitrary constants, while the degree is the highest power to which that derivative is raised once the equation is written as a polynomial in its derivatives. Starting with $y^{2}=a\left(x+\frac{\sqrt{a}}{2}\right)$, differentiating with respect to $x$ gives $2y \frac{dy}{dx} = a$. Substituting this expression for $a$ back into the original curve equation yields $y^2 = 2y \frac{dy}{dx} \left(x + \frac{\sqrt{2y \frac{dy}{dx}}}{2}\right)$, which simplifies to $y^2 - 2xy \frac{dy}{dx} = y \frac{dy}{dx} \sqrt{2y \frac{dy}{dx}}$. Squaring both sides to eliminate the radical produces the equation:

$(y^2 - 2xy \frac{dy}{dx})^2 = 8y^3 (\frac{dy}{dx})^3$

In this rationalized form, the highest derivative present is $\frac{dy}{dx}$, making the order 1, and the highest power to which this derivative is raised is 3, making the degree 3. Therefore, the difference between the degree and order is 2.

The differential equation
$\frac{dy}{dx}+2y\tan x=\sin x$
requires an integrating factor of $e^{\int 2\tan x \, dx} = \sec^2 x$, which transforms the expression into $\frac{d}{dx}(y \sec^2 x) = \sec x \tan x$. Integrating both sides leads to $y \sec^2 x = \sec x + C$, and applying the condition $y\left(\frac{\pi}{3}\right) = 0$ determines that $C = -2$, defining the function as $y = \cos x - 2\cos^2 x$. Completing the square to rewrite this as $y = -2\left(\cos x - \frac{1}{4}\right)^2 + \frac{1}{8}$ reveals that the maximum value of the function is $\frac{1}{8}$.

Separation of variables is the fundamental technique for solving differential equations of this form, allowing the terms involving $x$ and $y$ to be isolated on opposite sides for independent integration. Rearranging the expression yields

$\frac{y}{\sqrt{1 - y^{2}}} dy = -x e^{x} dx$

Integrating the left side with respect to $y$ results in $-\sqrt{1 - y^{2}}$, and integrating the right side by parts gives $-e^{x}(x - 1)$, establishing the relation

$\sqrt{1 - y^{2}} = e^{x}(x - 1) + C$

Using the initial condition $y(1) = -1$ determines that $C = 0$, simplifying the relation to $\sqrt{1 - y^{2}} = e^{x}(x - 1)$. Evaluating this at $x = 3$ yields

$\sqrt{1 - y^{2}} = e^{3}(3 - 1) = 2e^{3}$

Squaring both sides produces $1 - y^{2} = 4e^{6}$, which confirms the value of $(y(3))^{2}$ as $1 - 4e^{6}$.

Linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ are solved by identifying an integrating factor, $e^{\int P(x) dx}$, which allows the left side of the equation to be expressed as the derivative of a product. Rearranging the original equation $x^2 dy + (y - \frac{1}{x}) dx = 0$ results in $\frac{dy}{dx} + \frac{1}{x^2}y = \frac{1}{x^3}$, where $P(x) = \frac{1}{x^2}$. The integrating factor becomes $e^{\int \frac{1}{x^2} dx} = e^{-\frac{1}{x}}$.

Multiplying both sides of the rearranged equation by $e^{-\frac{1}{x}}$ allows for direct integration, yielding $y \cdot e^{-\frac{1}{x}} = \int \frac{1}{x^3} e^{-\frac{1}{x}} dx + c$. Using the substitution $t = -\frac{1}{x}$, where $dt = \frac{1}{x^2} dx$, the integral transforms into $\int -t e^t dt$. This evaluates to $-t e^t + e^t + c$, which translates back to $(\frac{1}{x} + 1)e^{-\frac{1}{x}} + c$.

Applying the initial condition $y(1) = 1$ gives $1 \cdot e^{-1} = (1 + 1)e^{-1} + c$, which simplifies to $c = -e^{-1}$. The resulting expression for the solution curve is $y = \frac{1}{x} + 1 - \frac{e^{\frac{1}{x}}}{e}$. Substituting $x = \frac{1}{2}$ into this expression gives $y = 2 + 1 - \frac{e^2}{e}$, resulting in the final value $3 - e$.

Substitution acts as the key to simplifying the differential equation $e^{\sin y} \cos y \frac{dy}{dx}+e^{\sin y} \cos x=\cos x$, allowing it to be treated as a first-order linear equation. By setting $t = e^{\sin y}$, the term $e^{\sin y} \cos y \frac{dy}{dx}$ becomes $\frac{dt}{dx}$, transforming the original expression into $\frac{dt}{dx} + t \cos x = \cos x$.

Multiplying this linear equation by the integrating factor $e^{\int \cos x \, dx} = e^{\sin x}$ leads to $\frac{d}{dx} (t e^{\sin x}) = \cos x e^{\sin x}$. Integrating both sides with respect to $x$ yields $t e^{\sin x} = e^{\sin x} + C$. Applying the condition $y(0) = 0$, where $t = e^{\sin 0} = 1$, reveals that $1 \cdot 1 = 1 + C$, so $C$ must be $0$.

Consequently, $t = e^{\sin y} = 1$, which implies $\sin y = 0$ and therefore $y = 0$ for all values of $x$. Substituting this result into the expression $1+y\left(\frac{\pi}{6}\right)+\frac{\sqrt{3}}{2}y\left(\frac{\pi}{3}\right)+\frac{1}{\sqrt{2}}y\left(\frac{\pi}{4}\right)$ results in $1 + 0 + 0 + 0$, which simplifies to $1$.

The rate at which a population changes over time follows the differential equation $\frac{dP}{dt} = 0.5P - 450$, which can be rearranged to $\frac{dP}{dt} = 0.5(P - 900)$ to separate the variables. Integrating $\frac{dP}{P - 900} = 0.5 dt$ leads to $\log_e |P - 900| = 0.5t + C$, where the initial condition $P(0) = 850$ identifies the constant as $C = \log_e 50$.

Setting $P = 0$ to find the time when the population disappears results in the equation $\log_e |-900| = 0.5t + \log_e 50$. This simplifies to $0.5t = \log_e 900 - \log_e 50$, or $0.5t = \log_e 18$, which gives the final value $t = 2 \log_e 18$.

To derive the differential equation for the system of parabolas defined by $y^2 = 4a(x+a)$, we must eliminate the arbitrary constant $a$ by establishing a relationship between the parameter and the derivative of the curve. First, expand the original equation to $y^2 = 4ax + 4a^2$ and differentiate both sides with respect to $x$, which results in $2y \frac{dy}{dx} = 4a$, or more simply, $a = \frac{y}{2} \frac{dy}{dx}$.

Substitute this expression for $a$ back into the original equation $y^2 = 4ax + 4a^2$. This substitution yields $y^2 = 4x \left( \frac{y}{2} \frac{dy}{dx} \right) + 4 \left( \frac{y}{2} \frac{dy}{dx} \right)^2$, which simplifies to $y^2 = 2xy \frac{dy}{dx} + y^2 \left( \frac{dy}{dx} \right)^2$. Dividing the entire resulting equation by $y$, assuming $y$ is non-zero, allows us to isolate the terms to reach the final form $y \left( \frac{dy}{dx} \right)^2 + 2x \frac{dy}{dx} - y = 0$.

Linear differential equations structured as $\frac{dy}{dx} + P(x)y = Q(x)$ are solved by utilizing an integrating factor calculated as $e^{\int P(x) \, dx}$. Given $P(x) = \tan x$, the integrating factor is $e^{\int \tan x \, dx}$, which simplifies to $\sec x$. Multiplying the differential equation by this factor transforms the left-hand side into the derivative of the product $y \sec x$, while the right-hand side becomes $\sec x \sin x$, equivalent to $\tan x$.

Integrating both sides with respect to $x$ results in $y \sec x = \int \tan x \, dx = \log_{e}(\sec x) + C$. Applying the initial condition $y(0) = 0$ determines the constant $C$ as $0$, since $\log_{e}(\sec 0) = \log_{e}(1) = 0$. The specific solution is then $y = \cos x \log_{e}(\sec x)$. Evaluating this at $x = \frac{\pi}{4}$ provides $y\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) \log_{e}\left(\sec \frac{\pi}{4}\right)$, which simplifies to $\frac{1}{\sqrt{2}} \log_{e}(\sqrt{2})$. Noting that $\log_{e}(\sqrt{2}) = \frac{1}{2} \log_{e} 2$, the final value is $\frac{1}{2\sqrt{2}} \log_{e} 2$.

The area bounded by a parabola and the $x$-axis between its roots $\alpha$ and $\beta$ is determined by the formula $\frac{|\beta-\alpha|^3}{6}$. First, integrating the differential equation $\frac{dy}{dx}=2(x+1)$ yields the parabola $y=x^{2+}2x+C$. The roots of this quadratic, where $y=0$, allow us to relate the coefficients and the interval length; specifically, the sum of roots is $\alpha+\beta = -2$ and the product is $\alpha\beta = C$, which leads to the interval length relation
$(\beta-\alpha)^2 = (\alpha+\beta)^{2-}4\alpha\beta = (-2)^{2-}4C = 4(1-C)$
This confirms that the interval length is $\beta-\alpha = 2\sqrt{1-C}$. Substituting this into the area formula gives
$\frac{(2\sqrt{1-C})^3}{6} = \frac{8(1-C)^{3/2}}{6} = \frac{4(1-C)^{3/2}}{3}$
Setting this expression equal to the given area value of $\frac{4\sqrt{8}}{3}$ implies $(1-C)^{3/2} = \sqrt{8}$, which simplifies to $1-C = 2$, resulting in $C = -1$. Substituting this value back into the curve equation yields $y=x^{2+}2x-1$, and evaluating at $x=1$ provides
$y(1) = 1^{2+}2(1)-1 = 2$

The slope of the curve $y=f(x)$ is given by the differential equation $\frac{dy}{dx} = \frac{2y}{x \ln x}$. Separating the variables yields $\frac{dy}{y} = \frac{2 dx}{x \ln x}$. Integrating both sides leads to $\ln |y| = 2 \ln |\ln x| + C$, where the right-hand integral is solved by substituting $u = \ln x$. Using logarithmic properties, this can be expressed as $\ln |y| = \ln ((\ln x)^2) + C$.

Substituting the point $(2, (\ln 2)^2)$ into this equation helps determine the constant of integration. Inserting $x = 2$ and $y = (\ln 2)^2$ gives $\ln ((\ln 2)^2) = \ln ((\ln 2)^2) + C$, confirming that $C = 0$. Consequently, the function simplifies to $y = (\ln x)^2$. Evaluating this result at $x = e$ gives $f(e) = (\ln e)^2$, which equals 1.

Separating the variables in the differential equation leads to $e^{-y} dy = e^{\alpha x} dx$, which integrates to $-e^{-y} = \frac{e^{\alpha x}}{\alpha} + C$. Inserting the boundary conditions $y(\ln 2) = \ln 2$ and $y(0) = \ln(\frac{1}{2})$ produces the respective relations $-\frac{1}{2} = \frac{2^{\alpha}}{\alpha} + C$ and $-2 = \frac{1}{\alpha} + C$. Subtracting these two results eliminates the constant $C$ to yield $\frac{3}{2} = \frac{2^{\alpha} - 1}{\alpha}$. Rearranging into the form $3\alpha = 2(2^{\alpha} - 1)$ and testing positive integers shows that $\alpha = 2$ is the solution, as $3(2) = 2(2^2 - 1)$ simplifies to the identity $6 = 6$.

The rate of bacterial growth is directly proportional to the current population, which is governed by the differential equation $\frac{dN}{dt} = \lambda N$. Solving this via separation of variables yields the exponential growth model $N(t) = N_0 e^{\lambda t}$, where $N_0$ represents the initial population of 1000.

When 2 hours have elapsed, the population increases by $20\%$ to reach 1200, which allows us to determine the growth constant. Substituting $t=2$ and $N=1200$ into the model gives $1200 = 1000 e^{2\lambda}$, simplifying to $1.2 = e^{2\lambda}$ or $e^{2\lambda} = 6/5$. By taking the natural logarithm, we find $2\lambda = \log_e(6/5)$, so $\lambda = \frac{1}{2} \log_e(6/5)$.

We are tasked with identifying the value associated with a population of 2000 at $t = \frac{k}{\log_e(6/5)}$. Placing this time value into our growth equation $N(t) = 1000 e^{\lambda t}$ yields $2000 = 1000 e^{\left( \frac{1}{2} \log_e(6/5) \right) \left( \frac{k}{\log_e(6/5)} \right)}$. Canceling the $\log_e(6/5)$ terms inside the exponent simplifies the expression to $2 = e^{k/2}$. Converting this into logarithmic form gives $k/2 = \log_e 2$, which implies $k = 2 \log_e 2$. Calculating the final expression $\left( \frac{k}{\log_e 2} \right)^2$ results in $\left( \frac{2 \log_e 2}{\log_e 2} \right)^2$, which simplifies to $2^2 = 4$.

Recognising that the equation can be rearranged into a linear form with $x$ as the dependent variable and $y$ as the independent variable is the key step here. By dividing the entire expression by $dy$ and then by $y$, we obtain the standard linear form $\frac{dx}{dy} + \frac{2}{y}x = 10y^2$. The integrating factor for this type of differential equation is found by evaluating $e^{\int \frac{2}{y} dy}$, which results in $y^2$. Multiplying the rearranged equation by this integrating factor allows the left side to be expressed as the derivative of the product $x y^2$, simplifying the integration process.

Integrating the right side, which becomes $10y^2 \cdot y^2 = 10y^4$, yields the general solution $x y^2 = 2y^5 + C$. Substituting the point $(0,1)$ into this general solution establishes that $0 \cdot 1^2 = 2(1)^5 + C$, confirming that the constant of integration $C$ is $-2$. With the specific solution $x y^2 = 2y^5 - 2$ now determined, substituting the point $(2, \beta)$ yields $2 \beta^2 = 2\beta^5 - 2$. Dividing this relationship by $2$ simplifies it to $\beta^5 - \beta^2 - 1 = 0$, meaning that $\beta$ must satisfy the polynomial equation $y^5 - y^2 - 1 = 0$.

Separating variables in the differential equation transforms $\frac{2+\sin x}{y+1} \cdot \frac{dy}{dx} = -\cos x$ into the following relation:

$\frac{1}{y+1} dy = \frac{-\cos x}{2+\sin x} dx$

Integrating both sides results in $\ln(y+1) = -\ln(2+\sin x) + C$, which simplifies to $(y+1)(2+\sin x) = K$. Using the initial condition $y(0)=1$ determines the constant $K = (1+1)(2+\sin 0) = 4$, yielding the solution function:

$y(x) = \frac{4}{2+\sin x} - 1$

Substituting $x=\pi$ gives $a = y(\pi) = \frac{4}{2+0} - 1 = 1$. The derivative $b = \left. \frac{dy}{dx} \right|_{x=\pi}$ is found by substituting $x=\pi$ and $y(\pi)=1$ into the original differential equation $\frac{dy}{dx} = \frac{-\cos x (y+1)}{2+\sin x}$, which results in $b = \frac{-(-1)(1+1)}{2+0} = 1$. The final ordered pair is $(1, 1)$.

Treating $x$ as a function of $y$ simplifies the differential equation $(y^2 - x) \frac{dy}{dx} = 1$, which can be rearranged into the standard linear form $\frac{dx}{dy} + x = y^2$. The integrating factor for this expression is $e^{\int 1 dy} = e^y$. Applying this factor leads to the product rule formulation $\frac{d}{dy}(x e^y) = y^2 e^y$, and integrating both sides produces $x e^y = \int y^2 e^y dy + C$.

Solving the integral on the right using integration by parts gives $x e^y = (y^2 - 2y + 2)e^y + C$. Plugging in the initial condition where $x=0$ and $y=1$ determines the constant, where $0 = (1^2 - 2(1) + 2)e^1 + C$, establishing that $C = -e$. Determining the x-intercept requires setting $y=0$, which yields $x e^0 = (0^2 - 2(0) + 2)e^0 - e$, simplifying to $x = 2 - e$.

Homogeneous differential equations are resolved by employing a substitution method to reduce them into a separable form. Rewriting the expression as $\frac{dy}{dx} = \frac{2xy + y^2}{2x^2}$ and applying the substitution $y = vx$ leads to the derivative $\frac{dy}{dx} = v + x\frac{dv}{dx}$. Placing these into the differential equation yields $v + x\frac{dv}{dx} = \frac{2x(vx) + (vx)^2}{2x^2}$, which simplifies to $v + x\frac{dv}{dx} = v + \frac{v^2}{2}$.

Subtracting $v$ from both sides leaves $x\frac{dv}{dx} = \frac{v^2}{2}$. Separating the variables allows for integration, $\int \frac{2}{v^2} dv = \int \frac{1}{x} dx$, yielding $-\frac{2}{v} = \ln x + C$. Reverting to the original variables with $v = \frac{y}{x}$ gives the expression $-\frac{2x}{y} = \ln x + C$. Utilizing the condition where $x = 1$ and $y = 2$ reveals that $-1 = \ln(1) + C$, confirming that $C = -1$.

The resulting function is defined by $-\frac{2x}{y} = \ln x - 1$, or equivalently $y = \frac{2x}{1 - \ln x}$. Calculating the value of $f\left(\frac{1}{2}\right)$ involves substituting $x = \frac{1}{2}$, which results in $f\left(\frac{1}{2}\right) = \frac{2(\frac{1}{2})}{1 - \ln\left(\frac{1}{2}\right)}$. Since $-\ln\left(\frac{1}{2}\right) = \log_e 2$, the expression simplifies to $\frac{1}{1 + \log_e 2}$.

The differential equation $\frac{dy}{dx} - \frac{y+3x}{\log_{e}(y+3x)} + 3 = 0$ simplifies into a directly separable form by identifying the repeating term $y+3x$ as a single variable. Defining $v = y+3x$ and differentiating with respect to $x$ gives $\frac{dv}{dx} = \frac{dy}{dx} + 3$, which allows for substitution into the equation to obtain $\frac{dv}{dx} = \frac{v}{\log_{e} v}$. Rearranging this leads to the separable expression:
$\frac{\log_{e} v}{v} \, dv = dx$
Integrating both sides with the substitution $z = \log_{e} v$ simplifies the left side to $\int z \, dz$, resulting in the relation $\frac{z^2}{2} = x + C$. Substituting $\log_{e}(y+3x)$ back for $z$ produces $\frac{1}{2} (\log_{e}(y+3x))^2 = x + C$, which rearranges to $x - \frac{1}{2} (\log_{e}(y+3x))^{2} = C$.

Solving this differential equation relies on recognizing it as a homogeneous equation of the form $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$, where the substitution $y=vx$ is the standard approach to separate variables. Replacing $y$ with $vx$ makes $\frac{dy}{dx} = v + x\frac{dv}{dx}$, transforming the original expression into $v + x\frac{dv}{dx} = \frac{vx^2}{x^2 + v^2x^2} = \frac{v}{1+v^2}$. Rearranging this leads to $x\frac{dv}{dx} = -\frac{v^3}{1+v^2}$, which integrates to $\int \frac{1+v^2}{v^3} dv = -\int \frac{dx}{x}$. Integrating both sides gives $-\frac{1}{2v^2} + \ln|v| = -\ln|x| + C$. Substituting $v = \frac{y}{x}$ back into the equation yields $-\frac{x^2}{2y^2} + \ln\left(\frac{y}{x}\right) = -\ln|x| + C$, which simplifies to $-\frac{x^2}{2y^2} + \ln y = C$. Applying the initial condition $y(1)=1$, we find that $-\frac{1}{2} + 0 = C$, so $C = -1/2$. Substituting these values into $-\frac{x^2}{2y^2} + \ln y = -1/2$ at the point where $y=e$ produces $-\frac{x^2}{2e^2} + 1 = -1/2$, which simplifies to $\frac{x^2}{2e^2} = 3/2$, leading directly to $x = \sqrt{3}e$.

Solving first-order differential equations often involves transforming the expression into a standard linear form to apply an integrating factor. Transforming the equation $e^y\left(\frac{dy}{dx}-1\right) = e^x$ requires the substitution $t = e^y$, which simplifies the expression to the linear equation $\frac{dt}{dx} - t = e^x$. Applying the integrating factor $e^{-x}$ results in $\frac{d}{dx}(t e^{-x}) = 1$, and integrating both sides yields $t e^{-x} = x + c$. Substituting $t = e^y$ and using the initial condition $y(0) = 0$ determines $c = 1$, leading to the relation $e^{y-x} = x + 1$. Finally, setting $x = 1$ in this expression results in $e^{y(1)-1} = 2$, which simplifies to $y(1) = 1 + \log_{e} 2$.

Separating the variables in the given differential equation leads to the form $\frac{dy}{\sqrt{1-y^2}} = -\frac{dx}{\sqrt{1-x^2}}$. Integrating both sides yields the relation $\sin^{-1}(y) = -\sin^{-1}(x) + C$. Using the initial condition $y\left(\frac{1}{2}\right) = \frac{\sqrt{3}}{2}$, we calculate $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = -\sin^{-1}\left(\frac{1}{2}\right) + C$, which simplifies to $\frac{\pi}{3} = -\frac{\pi}{6} + C$, providing the constant $C = \frac{\pi}{2}$. The specific solution is therefore $\sin^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(x)$. Substituting $x = -\frac{1}{\sqrt{2}}$ into this expression results in $\sin^{-1}(y) = \frac{\pi}{2} - \left(-\frac{\pi}{4}\right) = \frac{3\pi}{4}$, and taking the sine of both sides yields $y = \sin\left(\frac{3\pi}{4}\right) = \frac{1}{\sqrt{2}}$.