Differential Equations – JEE Main MathematicsPractice Questions & PYQs

The differential equation $x\sqrt{1-x^2} dy + (y\sqrt{1-x^2} - x \cos^{-1} x) dx = 0$ is a first-order linear differential equation. Rearranging the terms into the standard form $\frac{dy}{dx} + \frac{y}{x} = \frac{\cos^{-1} x}{\sqrt{1-x^2}}$ reveals the integrating factor is $e^{\int \frac{1}{x} dx} = x$. Multiplying the differential equation by this integrating factor allows the left side to be written as the derivative of the product $xy$, giving $xy = \int \frac{x \cos^{-1} x}{\sqrt{1-x^2}} dx + C$.

Evaluating the integral by substituting $u = \cos^{-1} x$ leads to $du = -\frac{1}{\sqrt{1-x^2}} dx$ and $x = \cos u$, which simplifies the integral to $-\int u \cos u du$. Applying integration by parts results in $-u \sin u - \cos u + C$, which corresponds to the expression $-(\cos^{-1} x)\sqrt{1-x^2} - x + C$. Applying the boundary condition $\lim_{x \to 1^-} y(x) = 1$ allows for the determination of the constant $C$ by substituting $x=1$, yielding $1(1) = 0 - 1 + C$, so $C = 2$. With the specific solution $xy = -(\cos^{-1} x)\sqrt{1-x^2} - x + 2$, evaluating at $x = \frac{1}{2}$ results in $\frac{1}{2} y\left(\frac{1}{2}\right) = -(\frac{\pi}{3})(\frac{\sqrt{3}}{2}) - \frac{1}{2} + 2$, which simplifies to $y\left(\frac{1}{2}\right) = 3 - \frac{\pi}{\sqrt{3}}$.

The total change in a quantity is determined by integrating its rate of change over the specified interval. In this context, calculating the net increase in production simply requires computing the definite integral of the marginal production function as the number of workers increases from zero to 25.

$\int_{0}^{25} (100 - 12\sqrt{x}) dx$

The antiderivative of the function is $100x - 8x^{3/2}$. Evaluating this expression at the upper limit of 25 yields $(100 \times 25) - (8 \times 25^{3/2}) = 2500 - 1000 = 1500$. By adding this net growth of 1500 items to the initial output of 2000 items, the new production level reaches 3500.

Integrating the slope $\frac{dy}{dx} = \frac{2 e^{2x} - 6 e^{-x} + 9}{2 + 9 e^{-2x}}$ requires simplifying the expression to $\frac{dy}{dx} = e^{2x} - \frac{6 e^{-x}}{2 + 9 e^{-2x}}$ and integrating term by term. The first part is straightforward, while the second part is solved by substituting $u = e^{-x}$ to perform an inverse tangent integration, yielding the general curve equation $y = \frac{e^{2x}}{2} + \sqrt{2} \tan^{-1}\left(\frac{3 e^{-x}}{\sqrt{2}}\right) + C$.

Substituting the given point $\left(0, \frac{1}{2} + \frac{\pi}{2\sqrt{2}}\right)$ into this equation allows for the determination of the constant $C = \frac{\pi}{2\sqrt{2}} - \sqrt{2} \tan^{-1}\left(\frac{3}{\sqrt{2}}\right)$. When the second point $\left(\alpha, \frac{1}{2} e^{2\alpha}\right)$ is applied, the term $\frac{e^{2\alpha}}{2}$ on both sides cancels out, leading to the equality $\tan^{-1}\left(\frac{3}{\sqrt{2}}\right) - \tan^{-1}\left(\frac{3 e^{-\alpha}}{\sqrt{2}}\right) = \frac{\pi}{4}$.

Applying the trigonometric identity $\tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$ with $A = \frac{3}{\sqrt{2}}$ and $B = \frac{3 e^{-\alpha}}{\sqrt{2}}$, the relation becomes $\frac{\frac{3}{\sqrt{2}} - \frac{3 e^{-\alpha}}{\sqrt{2}}}{1 + \frac{9}{2} e^{-\alpha}} = 1$. Solving this equation by clearing the fractions leads to $\frac{3}{\sqrt{2}} e^{\alpha} - \frac{3}{\sqrt{2}} = e^{\alpha} + \frac{9}{2}$, which simplifies directly to $e^{\alpha} = \frac{3}{\sqrt{2}} \left( \frac{3+\sqrt{2}}{3-\sqrt{2}} \right)$.

This problem involves solving a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, which is solved by finding an integrating factor to simplify the expression. By identifying $P(x) = \frac{4x}{x^{2-}1}$, we calculate the integrating factor as $e^{\int \frac{4x}{x^{2-}1} dx} = e^{2 \ln(x^{2-}1)} = (x^{2-}1)^2$. Multiplying the differential equation throughout by $(x^{2-}1)^2$ allows us to rewrite the left side as the derivative of the product $\frac{d}{dx}[y(x^{2-}1)^2]$, while the right side simplifies to $\frac{x+2}{\sqrt{x^{2-}1}}$.

Integrating both sides with respect to $x$ gives $y(x^{2-}1)^2 = \int \frac{x}{\sqrt{x^{2-}1}} dx + \int \frac{2}{\sqrt{x^{2-}1}} dx$. The first integral evaluates to $\sqrt{x^{2-}1}$ and the second to $2 \ln(x + \sqrt{x^{2-}1})$, resulting in the general solution $y(x^{2-}1)^2 = \sqrt{x^{2-}1} + 2 \ln(x + \sqrt{x^{2-}1}) + C$. Substituting the given condition $y(2) = \frac{2}{9} \ln(2+\sqrt{3})$ into this equation gives $9 \cdot \frac{2}{9} \ln(2+\sqrt{3}) = \sqrt{3} + 2 \ln(2+\sqrt{3}) + C$, which confirms that $C = -\sqrt{3}$.

With the constant determined, we evaluate the solution at $x = \sqrt{2}$ to find $y(\sqrt{2})((\sqrt{2})^{2-}1)^2 = \sqrt{2-1} + 2 \ln(\sqrt{2} + \sqrt{2-1}) - \sqrt{3}$, which simplifies to $1 + 2 \ln(\sqrt{2} + 1) - \sqrt{3}$. Comparing this expression to the required form $\alpha \log_e (\sqrt{\alpha}+\beta) + \beta - \sqrt{\gamma}$, we identify $\alpha = 2$, $\beta = 1$, and $\gamma = 3$. Multiplying these natural numbers together results in $\alpha \beta \gamma = 2 \times 1 \times 3 = 6$.

The differential equation $\frac{dy}{dx} + \frac{1}{x^{2}-1} y = \left( \frac{x-1}{x+1} \right)^{\frac{1}{2}}$ follows the standard linear first-order form $\frac{dy}{dx} + P(x)y = Q(x)$. The integrating factor, defined as $e^{\int P(x) dx}$, involves calculating $\int \frac{1}{x^{2-}1} dx$, which results in $\frac{1}{2} \ln \left( \frac{x-1}{x+1} \right)$. Consequently, the integrating factor simplifies to $\left( \frac{x-1}{x+1} \right)^{\frac{1}{2}}$. Multiplying the original equation by this integrating factor transforms the left side into the derivative of the product $y \cdot \left( \frac{x-1}{x+1} \right)^{\frac{1}{2}}$, while the right side becomes $\frac{x-1}{x+1}$. Integrating both sides gives $y \cdot \left( \frac{x-1}{x+1} \right)^{\frac{1}{2}} = \int \left( 1 - \frac{2}{x+1} \right) dx$, leading to the general solution $y \cdot \left( \frac{x-1}{x+1} \right)^{\frac{1}{2}} = x - 2 \ln(x+1) + C$.

Substituting the condition that the curve passes through the point $\left(2, \sqrt{\frac{1}{3}}\right)$ gives $\frac{1}{\sqrt{3}} \cdot \sqrt{\frac{1}{3}} = 2 - 2 \ln 3 + C$, which simplifies to $\frac{1}{3} = 2 - 2 \ln 3 + C$, and thus $C = 2 \ln 3 - \frac{5}{3}$. Applying this value to the solution at $x=8$ yields $y(8) \cdot \sqrt{\frac{8-1}{8+1}} = 8 - 2 \ln 9 + 2 \ln 3 - \frac{5}{3}$. Since $\sqrt{\frac{7}{9}} = \frac{\sqrt{7}}{3}$ and $2 \ln 9 = 4 \ln 3$, this simplifies to $y(8) \cdot \frac{\sqrt{7}}{3} = 8 - 4 \ln 3 + 2 \ln 3 - \frac{5}{3} = \frac{19}{3} - 2 \ln 3$. Multiplying both sides by 3 confirms that $\sqrt{7} y(8) = 19 - 6 \ln 3$.

The relationship between the slope of a curve and its coordinates defines a first-order separable differential equation where the rate of change $\frac{dy}{dx}$ is proportional to the ratio $-\frac{y}{x}$. Expressing this as $\frac{dy}{dx} = -k \frac{y}{x}$ and integrating both sides results in $\int \frac{dy}{y} = -k \int \frac{dx}{x}$, which leads to the general form $\ln |y| = -k \ln |x| + C$. Substituting the coordinates of the first point $(1, 2)$ determines the integration constant $C = \ln 2$, as $\ln 2 = -k \ln 1 + C$. Utilizing the second point $(8, 1)$ allows for finding the proportionality constant, where $\ln 1 = -k \ln 8 + \ln 2$ simplifies to $3k \ln 2 = \ln 2$, identifying that $k = \frac{1}{3}$. Evaluating the function at $x = \frac{1}{8}$ results in $\ln |y| = -\frac{1}{3} \ln \left( \frac{1}{8} \right) + \ln 2$. Because $-\frac{1}{3} \ln \left( \frac{1}{8} \right) = \ln 2$, the expression simplifies to $\ln |y| = \ln 2 + \ln 2 = 2 \ln 2$, confirming that $|y| = 4$.