Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Recognizing that the term $x dy + y dx$ represents the total differential $d(xy)$ transforms the differential equation $(1-x^2 y^2) dx = y dx + x dy$ into the separable form $dx = \frac{d(xy)}{1-(xy)^2}$. Integrating both sides yields the solution $x = \frac{1}{2} \ln \left| \frac{1+xy}{1-xy} \right| + C$. Applying the initial condition that $x=1$ when $y=2$ allows for the determination of the constant $C$, which is found by evaluating $1 = \frac{1}{2} \ln \left| \frac{1+2}{1-2} \right| + C$, resulting in $C = 1 - \frac{1}{2} \ln 3$. Substituting $x=2$ and $y=\alpha$ into the general solution produces $2 = \frac{1}{2} \ln \left| \frac{1+2\alpha}{1-2\alpha} \right| + 1 - \frac{1}{2} \ln 3$. Rearranging this equation to $2 + \ln 3 = \ln \left| \frac{1+2\alpha}{1-2\alpha} \right|$ and exponentiating both sides leads to the equation $3e^2 = \left| \frac{1+2\alpha}{1-2\alpha} \right|$. Solving for $\alpha$ using the negative case of the absolute value, $\frac{1+2\alpha}{1-2\alpha} = -3e^2$, provides the value $\alpha = \frac{1+3e^2}{2(3e^{2-}1)}$.

To solve a linear first-order differential equation, we first rearrange the equation into standard form to isolate the derivative term. Dividing the given expression $(x+1) y' - y = e^{3x}(x+1)^2$ by $(x+1)$ yields $y' - \frac{y}{x+1} = e^{3x}(x+1)$. We then identify the integrating factor, calculated as $e^{-\int \frac{1}{x+1} dx}$, which simplifies to $e^{-\ln(x+1)}$ or simply $\frac{1}{x+1}$. Multiplying the entire equation by this factor allows us to express the left side as the derivative of the product $y \cdot (\frac{1}{x+1})$, leading to $\frac{y}{x+1} = \int e^{3x} dx = \frac{e^{3x}}{3} + C$.

Using the provided initial condition $y(0) = \frac{1}{3}$, we can determine the constant of integration. Substituting these values into the equation results in $\frac{1/3}{1} = \frac{e^0}{3} + C$, which simplifies to $\frac{1}{3} = \frac{1}{3} + C$, confirming that $C=0$. The function is therefore defined by $y = \frac{e^{3x}(x+1)}{3}$. To determine the nature of the curve at $x=-\frac{4}{3}$, we examine the first derivative. Applying the product rule gives $y' = e^{3x}(x + \frac{4}{3})$, which shows that the derivative is zero at $x=-\frac{4}{3}$, confirming it is indeed a critical point.

We classify the nature of this critical point using the second derivative test. Differentiating $y'$ with respect to $x$ provides $y'' = 3e^{3x}(x + \frac{4}{3}) + e^{3x}$, which simplifies to $e^{3x}(3x+5)$. Evaluating this second derivative at the critical value $x=-\frac{4}{3}$ yields $y''(-\frac{4}{3}) = e^{3(-4/3)}(3(-\frac{4}{3}) + 5)$, which equals $e^{-4}(-4 + 5) = e^{-4}$. Since $e^{-4} > 0$, the function is concave up at this point, indicating that the curve has a local minimum at $x=-\frac{4}{3}$.

First-order linear differential equations are solved by identifying the integrating factor that simplifies the left side into the derivative of a product. Dividing the original equation by $(x-1)$ yields the standard form $\frac{dy}{dx} + \frac{2x}{x-1} y = \frac{1}{(x-1)^2}$, where the integrating factor is $e^{\int \frac{2x}{x-1} dx} = e^{2x + 2\ln(x-1)} = (x-1)^2 e^{2x}$. Multiplying the standard equation by this factor transforms it into \frac{d}{dx} \[y(x-1)^2 e^{2x}]$ = e^{2x}$. Integrating both sides with respect to$x$results in$y(x-1)^2 e^{2x} = \frac{1}{2} e^{2x} + C$.

Using the initial condition $y(2) = \frac{1+e^4}{2e^4}$ enables solving for the constant $C = \frac{1}{2}$, leading to the specific solution $y = \frac{e^{2x}+1}{2(x-1)^2 e^{2x}}$. Evaluating this function at $x=3$ yields $y(3) = \frac{e^{6+}1}{8e^6}$. Comparing this to the form $\frac{e^\alpha+1}{\beta e^\alpha}$ identifies $\alpha=6$ and $\beta=8$, meaning $\alpha+\beta = 14$.

Treating $y^2$ as an auxiliary variable allows us to simplify the differential equation $(x-y^{2}) d x + y(5 x + y^{2}) d y = 0$ into a more manageable form. Rearranging the terms leads to $y \frac{dy}{dx} = \frac{y^{2-}x}{5x+y^2}$, and by setting $t = y^2$ such that $\frac{1}{2} dt = y dy$, the equation becomes $\frac{1}{2} \frac{dt}{dx} = \frac{t-x}{5x+t}$. This substitution reveals a homogeneous structure, which is addressed by substituting $t = vx$ where $v$ is a function of $x$.

Upon substituting $t = vx$ and replacing the derivative with $v + x \frac{dv}{dx}$, we obtain $\frac{1}{2} (v + x \frac{dv}{dx}) = \frac{v-1}{5+v}$, which simplifies to $x \frac{dv}{dx} = \frac{-v^2 - 3v - 2}{5+v}$. Separating the variables gives $\int \frac{v+5}{v^{2+}3v+2} dv = -\int \frac{dx}{x}$. Utilizing partial fraction decomposition on the integrand $\frac{v+5}{(v+1)(v+2)}$ yields $\int (\frac{4}{v+1} - \frac{3}{v+2}) dv = -\int \frac{dx}{x}$.

Integrating both sides leads to $4 \ln|v+1| - 3 \ln|v+2| = -\ln|x| + C$. Substituting $v = \frac{y^2}{x}$ back into the expression results in $\ln \left| \frac{(\frac{y^2}{x}+1)^4}{(\frac{y^2}{x}+2)^3} \right| = -\ln|x| + C$. Applying logarithmic properties and simplifying the fraction gives $\frac{(y^{2+}x)^4}{x(y^{2+}2x)^3} = \frac{C}{x}$, which reduces directly to $(y^{2+}x)^4 = C \left| (y^{2+}2x)^3 \right|$.

This problem relies on identifying a first-order linear differential equation, which can be solved by isolating the terms involving $y$ and the derivative. Rewriting the expression $\frac{dy}{dx} = 2\tan x(\cos x - y)$ in the standard form $\frac{dy}{dx} + (2\tan x)y = 2\sin x$ reveals the necessary structure to determine the integrating factor. The integrating factor is found by exponentiating the integral of the coefficient of $y$, which is $e^{\int 2\tan x\,dx} = e^{2\ln|\sec x|} = \sec^2 x$.

Multiplying the entire differential equation by $\sec^2 x$ allows the left side to be expressed as the derivative of the product $y\sec^2 x$, simplifying the right side to $\int 2\sin x \sec^2 x\,dx = \int 2\tan x \sec x\,dx = 2\sec x + c$. Substituting the provided coordinates $\left(\frac{\pi}{4}, 0\right)$ into this solution leads to $0 = 2\sqrt{2} + c$, which establishes that the constant $c$ is $-2\sqrt{2}$. Rearranging the resulting equation provides the explicit function $y = 2\cos x - 2\sqrt{2}\cos^2 x$.

The final value is obtained by computing the definite integral $\int_{0}^{\pi/2} (2\cos x - 2\sqrt{2}\cos^2 x) \, dx$. Evaluating the first term, $\int_{0}^{\pi/2} 2\cos x \, dx$, yields $2\sin x$ evaluated from $0$ to $\frac{\pi}{2}$, resulting in $2$. The second term is resolved using the trigonometric identity $\cos^2 x = \frac{1+\cos 2x}{2}$, transforming $-2\sqrt{2}\int_{0}^{\pi/2} \cos^2 x \, dx$ into $-\sqrt{2} \int_{0}^{\pi/2} (1+\cos 2x) \, dx$, which evaluates to $-\sqrt{2}(\frac{\pi}{2}) = -\frac{\pi}{\sqrt{2}}$. Combining these results gives $2 - \frac{\pi}{\sqrt{2}}$.

For the differential equation $\frac{dy}{dx} = \frac{a x - b y + a}{b x + c y + a}$ to represent a circle, the variables must satisfy specific symmetry conditions that allow the expression to be integrated into an equation of a conic section. By rearranging the terms and integrating, the equation takes the form $\frac{c y^{2}}{2} + a y - \frac{a x^{2}}{2} - a x + b x y = k$. For this curve to be a circle, the coefficients of the squared terms must be identical and the $xy$ term must be eliminated, which necessitates that $a = -c$ and $b = 0$.

Substituting these conditions back into the general form leads to $x^{2} + y^{2} + 2 x - 2 y = \lambda$. Given that the circle passes through the point $(2,5)$, we substitute these coordinates to find $\lambda$, yielding $4 + 25 + 4 - 10 = \lambda$, which results in $\lambda = 23$. The equation of the circle is thus $x^{2} + y^{2} + 2 x - 2 y - 23 = 0$. Completing the square for $x$ and $y$ reveals the center at $(-1, 1)$ and a radius of $\sqrt{(-1)^{2} + 1^{2} + 23} = 5$. The distance from the center $(-1, 1)$ to the target point $(11,6)$ is calculated using the distance formula, resulting in $\sqrt{(11 - (-1))^{2} + (6 - 1)^{2}} = \sqrt{12^{2} + 5^{2}} = 13$. Subtracting the radius from this total distance gives a final shortest distance of 8.

Solving a linear first-order differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$ begins by normalizing the original equation by dividing it by $\sin^2 2x$. This transformation simplifies the coefficient of $y$ because $\frac{2\sin 4x}{\sin^2 2x}$ becomes $4\cot 2x$ using the identity $\sin 4x = 2\sin 2x \cos 2x$. Consequently, the equation takes the standard form $\frac{dy}{dx} + (8 + 4\cot 2x)y = 2 e^{-4x}\frac{2\sin 2x + \cos 2x}{\sin^2 2x}$.

The integrating factor for this equation is determined by $e^{\int (8 + 4\cot 2x) \, dx}$, which yields $e^{8x + 2\ln(\sin 2x)}$ or equivalently $e^{8x} \sin^2 2x$. Multiplying the entire differential equation by this factor allows the left side to be written as the derivative of the product $y \cdot e^{8x} \sin^2 2x$. Integrating the right side involves computing $\int 2 e^{4x}(2\sin 2x + \cos 2x) \, dx$, which results in $e^{4x} \sin 2x + c$ because the integrand is the exact derivative of $e^{4x} \sin 2x$.

Applying the initial condition $y(\pi/4) = e^{-\pi}$ to the equation $y \cdot e^{8x} \sin^2 2x = e^{4x} \sin 2x + c$ allows us to find the constant $c$. Substituting these values leads to $e^\pi = e^\pi + c$, which confirms that $c = 0$. The simplified general solution for the curve is $y = \frac{1}{e^{4x} \sin 2x}$. Evaluating this function at $x = \pi/6$ gives $\frac{1}{e^{4(\pi/6)} \sin(\pi/3)}$, which simplifies to $\frac{2}{\sqrt{3}} e^{-2\pi/3}$.

A first-order linear differential equation takes the form $\frac{dy}{dx} + P(x)y = Q(x)$, which can be solved systematically by finding an integrating factor. Rearranging the expression $x(1-x^2)\frac{dy}{dx} + (3x^2 y - y - 4x^3) = 0$ yields $\frac{dy}{dx} + \frac{3x^{2-}1}{x(1-x^2)}y = \frac{4x^3}{x(1-x^2)}$, simplified to $\frac{dy}{dx} + \frac{3x^{2-}1}{x-x^3}y = \frac{4x^2}{1-x^2}$ for $x>1$. The integrating factor $IF$ is $e^{\int \frac{3x^{2-}1}{x-x^3} dx}$, which evaluates to $e^{-\ln|x^{3-}x|}$, or $\frac{1}{x^{3-}x}$.

Multiplying the differential equation by this integrating factor transforms it into the derivative of the product $\frac{d}{dx} \left( \frac{y}{x^{3-}x} \right) = \frac{4x^2}{1-x^2} \cdot \frac{1}{x^{3-}x}$. This simplifies to the integral $\int \frac{-4x}{(x^{2-}1)^2} dx$, resulting in $\frac{y}{x^{3-}x} = \frac{2}{x^{2-}1} + c$.

Applying the condition $y(2)=-2$ determines the constant: $\frac{-2}{8-2} = \frac{2}{4-1} + c$, which simplifies to $-\frac{1}{3} = \frac{2}{3} + c$, so $c = -1$. With the constant found, the equation becomes $\frac{y}{x^{3-}x} = \frac{2}{x^{2-}1} - 1$. Evaluating at $x=3$ gives $\frac{y}{27-3} = \frac{2}{9-1} - 1$, reducing to $\frac{y}{24} = \frac{2}{8} - 1$, or $\frac{y}{24} = -\frac{3}{4}$, resulting in $y = -18$.

The linear differential equation $\frac{dy}{dx} - y = 2 - e^{-x}$ has an integrating factor of $e^{-x}$. Multiplying by this factor and integrating provides the general solution $y = -2 + e^{-x} + Ce^x$. Because the condition $\lim_{x\to \infty } y(x)$ must be finite, the constant $C$ must be $0$, defining the specific curve as $y = -2 + e^{-x}$. At $x=0$, the function value is $y = -1$ and the derivative $\frac{dy}{dx} = -e^{-x}$ evaluates to a slope of $-1$, making the tangent line equation $y - (-1) = -1(x - 0)$, which simplifies to $x + y = -1$. The $x$-intercept $a$ and $y$-intercept $b$ are both $-1$, so the expression $a - 4b$ calculates to $-1 - 4(-1) = 3$.

This problem requires solving a first-order linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which relies on calculating the integrating factor $I.F. = e^{\int P(x) dx}$ to simplify the differential equation into the derivative of a product.

To determine the integrating factor, factor the denominator of the coefficient function $P(x)$, noting that $x^{3+}6x^{2+}11x+6$ is equivalent to $(x+1)(x+2)(x+3)$. Using partial fraction decomposition, the expression $\frac{2x^{2+}11x+13}{(x+1)(x+2)(x+3)}$ can be rewritten as $\frac{2}{x+1} + \frac{1}{x+2} - \frac{1}{x+3}$. Integrating these individual terms yields $2\ln(x+1) + \ln(x+2) - \ln(x+3)$, leading to an integrating factor of $\frac{(x+1)^2(x+2)}{x+3}$.

Multiplying the entire differential equation by this integrating factor makes the left side the derivative of $y \cdot \frac{(x+1)^2(x+2)}{x+3}$, so the solution is found by integrating the right side: $\int \frac{x+3}{x+1} \cdot \frac{(x+1)^2(x+2)}{x+3} dx$. This simplifies to the integral of $(x+1)(x+2)$, which is $\int (x^{2+}3x+2) dx = \frac{x^3}{3} + \frac{3x^2}{2} + 2x + c$. Applying the initial condition $(0,1)$, we find $1 \cdot \frac{1^2 \cdot 2}{3} = 0 + c$, giving $c = \frac{2}{3}$. Finally, substituting $x=1$ into the general solution gives $y(1) \cdot \frac{2^2 \cdot 3}{4} = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3}$. Simplifying the left side to $3y(1)$ and summing the right side to $\frac{9}{2}$ results in $y(1) = \frac{3}{2}$.

The differential equation $\frac{dy}{dx} = \frac{ax - by + a}{bx + cy + a}$ is rearranged by cross-multiplying to obtain $b(x dy + y dx) + (cy + a) dy = (ax + a) dx$. Noting that $x dy + y dx$ represents the differential of $xy$, the equation can be integrated on both sides to yield $b xy + \frac{cy^2}{2} + ay = \frac{ax^2}{2} + ax + k$. Rearranging these terms into the standard form of a general quadratic equation results in $\frac{ax^2}{2} - \frac{cy^2}{2} - b xy + ax - ay + k = 0$.

For this expression to represent a circle, the coefficient of the $xy$ term must be zero, which implies $b = 0$. Furthermore, the coefficients of the $x^2$ and $y^2$ terms must be equal, leading to the condition $\frac{a}{2} = -\frac{c}{2}$, or $c = -a$. Substituting $b = 0$ and $c = -a$ into the quadratic equation provides $\frac{ax^2}{2} + \frac{ay^2}{2} + ax - ay + k = 0$. Dividing the entire equation by $\frac{a}{2}$ simplifies the expression to $x^2 + y^2 + 2x - 2y + \frac{2k}{a} = 0$.

Identifying this with the general circle equation $x^2 + y^2 + 2gx + 2fy + c = 0$, the center coordinates are determined by $(-g, -f)$. Since $2g = 2$ and $2f = -2$, the center is located at $(-1, 1)$, meaning $\alpha = -1$ and $\beta = 1$. Consequently, the value of $\alpha + 2\beta$ is calculated as $-1 + 2(1) = 1$.

First-order linear differential equations follow the structure $\frac{dy}{dx} + P(x)y = Q(x)$ and are solved by finding an integrating factor, $e^{\int P(x) \, dx}$, to condense the left side into a product rule derivative. Dividing the given expression $\left(1+e^{2x}\right)\left(\frac{dy}{dx}+y\right)=1$ by the coefficient $\left(1+e^{2x}\right)$ reveals the standard form:
$\frac{dy}{dx} + y = \frac{1}{1+e^{2x}}$
The integrating factor is $e^{\int 1 \, dx} = e^x$, and multiplying the equation by this factor allows the left side to be expressed as the derivative of the product $e^x y(x)$. Integrating both sides with respect to $x$ yields:
$e^x y(x) = \int \frac{e^x}{1+e^{2x}} \, dx$
Evaluating this integral using the substitution $u = e^x$ results in $e^x y(x) = \arctan(e^x) + C$. Substituting the condition that the curve passes through the point $\left(0, \frac{\pi}{2}\right)$ gives $\frac{\pi}{2} = \arctan(1) + C$, which simplifies to $C = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$. With the constant determined, the limit as $x$ approaches infinity for the expression $e^x y(x) = \arctan(e^x) + \frac{\pi}{4}$ relies on the horizontal asymptote of the arctangent function, which is $\frac{\pi}{2}$, resulting in a final value of $\frac{3\pi}{4}$.

Treating $x$ as the dependent variable and $y$ as the independent variable reveals that the equation $(\tan^{-1} y - x) dy = (1+y^2) dx$ fits the structure of a first-order linear differential equation. Dividing the entire expression by $dy(1+y^2)$ transforms the equation into the standard form
$\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{\tan^{-1} y}{1+y^2}$
The integrating factor for this differential equation is $e^{\int \frac{1}{1+y^2} dy} = e^{\tan^{-1} y}$. Multiplying the equation by this factor allows the left side to be expressed as the derivative of the product $x e^{\tan^{-1} y}$, resulting in the integral equation
$x e^{\tan^{-1} y} = \int \frac{\tan^{-1} y}{1+y^2} e^{\tan^{-1} y} dy + C$
Substitution of $u = \tan^{-1} y$ turns the integral on the right into $\int u e^u du$, which evaluates to $u e^u - e^u + C$ using integration by parts. Substituting back for $u$, the general solution becomes $x e^{\tan^{-1} y} = e^{\tan^{-1} y} (\tan^{-1} y - 1) + C$. Applying the condition that the curve passes through the point $(1, 0)$ requires setting $x=1$ and $y=0$, leading to $1 \cdot e^0 = e^0(0 - 1) + C$. This simplifies to $1 = -1 + C$, which fixes the constant at $C = 2$. Finding the abscissa $x$ when the ordinate $y = \tan(1)$ involves setting $\tan^{-1} y = 1$ in the specific solution, yielding $x e = e(1 - 1) + 2$, which simplifies to $x e = 2$, or $x = \frac{2}{e}$.

The differential equation $\frac{dy}{dx} = x+y$ can be solved by using the substitution $t = x+y$, which transforms the expression into $\frac{dt}{dx} - 1 = t$, or simply $\frac{dt}{dx} = t+1$. Integrating this separable equation yields the general relation $x + y + 1 = Ce^x$, where $C$ is a constant determined by the specific initial conditions provided. For the first solution $y_1(x)$ satisfying $y_1(0)=0$, substituting these values gives $0+0+1 = Ce^0$, which identifies $C=1$. Similarly, for the second solution $y_2(x)$ satisfying $y_2(0)=1$, substituting these values into the general relation results in $0+1+1 = Ce^0$, which identifies $C=2$.

Equating the two resulting expressions for $y$ to find intersection points requires that $x + y_1 + 1 = x + y_2 + 1$, which simplifies to $e^x = 2e^x$. Since $e^x$ is strictly positive and can never be zero, dividing both sides by $e^x$ results in the impossible statement $1 = 2$. Because there are no values of $x$ that satisfy this equality, the two curves never intersect.

This differential equation exhibits the structure of a homogeneous equation, which can be identified by the ratio $\frac{y}{x}$. Rearranging the initial expression, we obtain the derivative:
$\frac{dy}{dx} = \frac{y}{x} + \sqrt{\left( \frac{y}{x} \right)^2 + 16}$
Substituting $v = \frac{y}{x}$ yields $y = vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Replacing $y$ in the differential equation leads to:
$v + x \frac{dv}{dx} = v + \sqrt{v^2 + 16} \implies x \frac{dv}{dx} = \sqrt{v^2 + 16}$
Separating variables and integrating both sides results in the following:
$\int \frac{dv}{\sqrt{v^2 + 16}} = \int \frac{dx}{x} \implies \ln |v + \sqrt{v^2 + 16}| = \ln |x| + C$
Applying the properties of logarithms and re-substituting $v = \frac{y}{x}$ yields the relationship:
$v + \sqrt{v^2 + 16} = Cx \implies y + \sqrt{y^2 + 16x^2} = Cx^2$
Using the initial condition $y(1) = 3$, we substitute $x = 1$ and $y = 3$ to find the constant:
$3 + \sqrt{3^2 + 16(1)^2} = C(1)^2 \implies 3 + 5 = C \implies C = 8$
With the constant determined, we evaluate $y$ at $x = 2$:
$y + \sqrt{y^2 + 16(2)^2} = 8(2)^2 \implies y + \sqrt{y^2 + 64} = 32$
Isolating the square root and squaring both sides gives $y^2 + 64 = (32 - y)^2$, which simplifies to $y^2 + 64 = 1024 - 64y + y^2$. Solving the resulting equation $64y = 960$ provides the final value of $y = 15$.