Differential Equations – JEE Main MathematicsPractice Questions & PYQs

Eliminating the arbitrary constants $c_1$ and $c_2$ from the expression $y = c_1 e^{c_2 x}$ relies on differentiating the function and observing patterns within the derivatives. The first derivative with respect to $x$ is $y' = c_1 c_2 e^{c_2 x}$, which can be rewritten as $y' = c_2 y$ by recognizing the original function $y$ within the expression. Differentiating $y' = c_2 y$ with respect to $x$ again yields $y'' = c_2 y'$. By substituting the value $c_2 = \frac{y'}{y}$ obtained from the first derivative into this second derivative equation, the constants are eliminated to form $y'' = \left(\frac{y'}{y}\right) y'$, which simplifies to $y y'' = (y')^2$.

This homogeneous differential equation can be simplified by substituting $y=vx$, which implies $\frac{dy}{dx} = v + x \frac{dv}{dx}$. Replacing the terms in the original equation yields $v + x \frac{dv}{dx} = 1 + v$, which simplifies directly to $x \frac{dv}{dx} = 1$. Rearranging this to $\frac{dv}{dx} = \frac{1}{x}$ allows us to integrate both sides, giving $\int dv = \int \frac{1}{x} dx$, which results in $v = \ln x + c$. Substituting $v = \frac{y}{x}$ back into the expression gives the general solution $y = x \ln x + cx$. Applying the initial condition $y(1) = 1$ leads to $1 = 1 \ln(1) + c(1)$, confirming that $c=1$ and resulting in the specific solution $y = x \ln x + x$.

Any circle passing through the origin with its center located on the $x$-axis follows the general equation $x^2 + y^2 + 2gx = 0$, where $g$ is a parameter that determines the circle's radius and center position. Because this family of curves relies on a single arbitrary constant, we eliminate it by differentiating the equation with respect to $x$ to uncover the underlying differential relationship.

Differentiating $x^2 + y^2 + 2gx = 0$ with respect to $x$ yields $2x + 2y \frac{dy}{dx} + 2g = 0$. From this expression, isolating the constant $g$ gives $g = -(x + y \frac{dy}{dx})$. Substituting this value back into the original equation results in $x^2 + y^2 + 2(-(x + y \frac{dy}{dx}))x = 0$. Simplifying the algebraic terms leads to $x^2 + y^2 - 2x^2 - 2xy \frac{dy}{dx} = 0$, which rearranges to the final form $y^2 = x^2 + 2xy \frac{dy}{dx}$.

The order of a differential equation is determined by the number of arbitrary constants to be eliminated, so the presence of two constants, $A$ and $B$, inherently requires two successive differentiations. Differentiation of $Ax^{2+}By^2=1$ with respect to $x$ yields $Ax + By \frac{dy}{dx} = 0$. Performing a second differentiation produces $A + By \frac{d^2 y}{d x^2} + B \left( \frac{dy}{dx} \right)^2 = 0$. By algebraically eliminating the constants, the final expression contains the second derivative $\frac{d^2y}{dx^2}$ raised to the power of one, which confirms the differential equation is of second order and first degree.

Homogeneous differential equations are effectively solved using the substitution $y = vx$, which transforms the expression into a separable form suitable for integration. Substituting $y = vx$ and the derivative $\frac{dy}{dx} = v + x \frac{dv}{dx}$ into the equation $x \frac{dy}{dx} = y(\log y - \log x + 1)$ simplifies the relationship to $v + x \frac{dv}{dx} = v(\log v + 1)$. Subtracting $v$ from both sides leaves $x \frac{dv}{dx} = v \log v$, which rearranges into the separable form $\frac{dv}{v \log v} = \frac{dx}{x}$. Integrating both sides involves the substitution $\log v = z$, which leads to $\log(\log v) = \log x + \log c$ and simplifies to $\log v = cx$. Replacing $v$ with its original definition of $\frac{y}{x}$ produces the final solution $\log \left( \frac{y}{x} \right) = cx$.

Eliminating the parameter from a family of curves requires differentiating the given relation with respect to $x$ and substituting the result back into the original form. Applying this to $y^2=2c\left(x+\sqrt{c}\right)$ yields $2yy'=2c$, which simplifies to $c=yy'$. Substituting this into the initial equation gives $y^2=2yy'\left(x+\sqrt{yy'}\right)$, and dividing by $y$ results in $y=2xy'+2y'\sqrt{yy'}$. Rearranging the terms to isolate the radical and squaring both sides produces the rationalized differential equation
$\left(y-2xy'\right)^2=4y\left(y'\right)^3$
Since the highest derivative present is the first derivative $y'$, the order is 1, and because the highest power of this derivative in the resulting polynomial form is 3, the degree is 3.

Differentiating the circle equation $x^{2+}y^{2-}2ay=0$ implicitly with respect to $x$ yields $2x + 2yy' - 2ay' = 0$, which allows the arbitrary constant $a$ to be expressed as $a = \frac{x+yy'}{y'}$. Inserting this value for $a$ back into the original equation results in $x^2 + y^2 - 2y \left( \frac{x+yy'}{y'} \right) = 0$. After multiplying through by $y'$ to clear the denominator, the expression becomes $(x^{2+}y^2)y' - 2xy - 2y^2y' = 0$. Collecting the $y'$ terms leads to $(x^{2-}y^2)y'=2xy$.

This problem presents a Bernoulli differential equation, which can be transformed into a linear form through an appropriate substitution. Beginning with the expression $y dx + (x + x^2 y) dy = 0$, we first rearrange the terms to isolate the derivative, resulting in $\frac{dx}{dy} + \frac{x}{y} = -x^2$. Dividing the entire equation by $x^2$ allows us to define the equation in terms of $x^{-2} \frac{dx}{dy} + x^{-1} \frac{1}{y} = -1$.

By introducing the substitution $t = x^{-1}$, we find that $\frac{dt}{dy} = -x^{-2} \frac{dx}{dy}$, which transforms the equation into the linear first-order differential form $\frac{dt}{dy} - \frac{t}{y} = 1$. Determining the integrating factor requires calculating $e^{\int -\frac{1}{y} dy}$, which simplifies to $e^{-\log y} = \frac{1}{y}$. Multiplying this factor across the linear equation and integrating provides the relation $t \cdot \frac{1}{y} = \log y + C$. Substituting $t = \frac{1}{x}$ back into this result yields $\frac{1}{xy} = \log y + C$, which rearranges to the final solution $-\frac{1}{xy} + \log y = C$.

The original equation can be rearranged by recognizing that $\frac{dy}{dx} = \frac{1}{dx/dy}$, which transforms the given expression into $(1+y^2) \frac{dx}{dy} + x = e^{\tan^{-1} y}$. Dividing through by $(1+y^2)$ brings this into the linear form $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{e^{\tan^{-1} y}}{1+y^2}$, where the coefficient of $x$ is $P(y) = \frac{1}{1+y^2}$. The integrating factor is determined by evaluating $e^{\int P(y) dy}$, which results in $e^{\tan^{-1} y}$. Multiplying the linear equation by this factor allows the left side to be expressed as the derivative of the product $x e^{\tan^{-1} y}$, while the right side becomes the integral $\int \frac{e^{2\tan^{-1} y}}{1+y^2} dy$. Performing this integration using the substitution $u = \tan^{-1} y$ simplifies the expression to $\frac{e^{2u}}{2} + C$, which is equivalent to $\frac{e^{2\tan^{-1} y}}{2} + C$. Equating the two sides and multiplying by 2 clears the fraction, resulting in $2x e^{\tan^{-1} y}=e^{2\tan^{-1} y}+k$.

The order of a differential equation corresponds to the number of independent arbitrary constants in the family of curves, as each constant must be eliminated through successive differentiation. For a parabola with its axis along the x-axis defined by $y^2 = 4a(x-h)$, we must eliminate the two constants $a$ and $h$. Differentiating once gives $2yy_1 = 4a$, or $yy_1 = 2a$, and differentiating again yields $y_1^2 + yy_2 = 0$. Because the highest derivative is the second derivative and its power is one, the degree and order are one and two, respectively.

Integrating a second-order differential equation requires finding the antiderivative twice in succession to isolate the dependent variable. Applying this process to $\frac{d^2 y}{d x^2}=e^{-2 x}$, the first integration with respect to $x$ yields the first derivative $\frac{dy}{dx}=\frac{e^{-2 x}}{-2}+c$, where $c$ serves as the initial constant of integration. A second integration of this result with respect to $x$ produces the final expression, where the term $\frac{e^{-2 x}}{-2}$ becomes $\frac{e^{-2 x}}{4}$, the constant $c$ integrates to $cx$, and the introduction of a new integration constant $d$ completes the solution as $y = \frac{e^{-2 x}}{4} + cx + d$.

The order of a differential equation is defined by the highest derivative present in the expression, while the degree corresponds to the exponent of that highest derivative once the equation is rewritten as a polynomial. For the equation $\left(1+3\; \frac{dy}{dx}\right)^{\frac{2}{3}} = 4\; \frac{d^3 y}{dx^3}$, the highest derivative is the third-order derivative, $\frac{d^3 y}{dx^3}$, which establishes the order as 3. By cubing both sides to eliminate the fractional exponent, the equation becomes $\left(1+3\; \frac{dy}{dx}\right)^2 = (4)^3 \left(\frac{d^3 y}{dx^3}\right)^3$, confirming that the highest-order derivative is raised to the power of 3 and the degree is 3.

A first-order linear differential equation, characterized by the form $\frac{dy}{dx} + P(x)y = Q(x)$, is resolved by applying an integrating factor $I.F. = e^{\int P(x) dx}$. This factor effectively transforms the left side into a total derivative, enabling integration and the determination of the constant using the boundary condition provided at the origin. Exploiting symmetry within the integration limits further simplifies the calculation of the definite integral for such functions.

In this specific equation, $P(x) = \frac{x}{x^{2-}1}$, which leads to the integrating factor $\sqrt{1-x^2}$ for the range $x \in (-1,1)$. Multiplying the differential equation by this factor expresses the left side as the derivative of $y\sqrt{1-x^2}$, which integrates to $\frac{x^5}{5} + x^2 + c$. Since the curve passes through the origin, $c$ must be zero, resulting in the function $f(x) = \frac{x^{5+}5x^2}{5\sqrt{1-x^2}}$.

Because $\frac{x^5}{5\sqrt{1-x^2}}$ is an odd function, its integral over the symmetric interval $[-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2}]$ is zero, leaving only the contribution of the even part, which simplifies to $2 \int_{0}^{\frac{\sqrt{3}}{2}} \frac{x^2}{\sqrt{1-x^2}} dx$. Substituting $x = \sin\theta$ changes the integral into $\int_{0}^{\frac{\pi}{3}} (1 - \cos 2\theta) d\theta$. Evaluating this at the limits $0$ and $\frac{\pi}{3}$ yields $\frac{\pi}{3} - \frac{\sqrt{3}}{4}$.

Linear differential equations of the form $\frac{dy}{dx} + P(x)y = Q(x)$ are solved by finding an integrating factor, defined as $I.F. = e^{\int P(x) dx}$. For the given equation, identifying $P(x)$ as $\frac{-3 x^5 \tan^{-1}(x^3)}{(1+x^6)^{3/2}}$ allows us to evaluate the integral, which results in the integrating factor $e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}}$.

Multiplying the entire differential equation by this integrating factor allows the left side to be expressed as the derivative of the product $y \cdot e^{\frac{\tan^{-1}(x^3) - x^3}{\sqrt{1+x^6}}}$. On the right side, the product of $2x \exp\left(\frac{x^{3-}\tan^{-1}(x^3)}{\sqrt{1+x^6}}\right)$ and the integrating factor cancels the exponential terms entirely, simplifying the integral to $\int 2x dx$. Integrating this yields $x^2 + c$. Since the solution passes through the origin $(0,0)$, substituting these coordinates confirms that $c = 0$, giving the relationship $y = x^2 e^{\frac{x^3 - \tan^{-1}(x^3)}{\sqrt{1+x^6}}}$. Evaluating this at $x=1$ provides $y(1) = 1^2 \cdot e^{\frac{1 - \pi/4}{\sqrt{2}}}$, which simplifies to $\exp\left(\frac{4-\pi}{4\sqrt{2}}\right)$.

Transforming the given differential equation into the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$ facilitates a straightforward integration using an integrating factor. Rearranging the original expression and rationalizing the right-hand side yields the simplified linear equation $\frac{dy}{dx} + \frac{y}{x} = x + \sqrt{x^2 - 1}$. Multiplying this by the integrating factor $x$ results in the derivative of the product $xy$, and integrating both sides with respect to $x$ gives $xy = \frac{x^3}{3} + \frac{(x^2 - 1)^{3/2}}{3} + C$. Applying the initial condition $y(1) = 1$ leads to $1 = \frac{1}{3} + 0 + C$, which identifies the constant $C = \frac{2}{3}$. Substituting $x = \sqrt{5}$ into the expression provides $\sqrt{5}y = \frac{(\sqrt{5})^3 + (5 - 1)^{3/2} + 2}{3}$, simplifying to $y = \frac{5\sqrt{5} + 10}{3\sqrt{5}}$. Since this value is approximately $3.156$, the greatest integer less than $y(\sqrt{5})$ is $3$.

Differential equations involving the ratio $\frac{y}{x}$ are classified as homogeneous, which simplifies the process by substituting $v = \frac{y}{x}$. Setting $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$ transforms the given equation into a simpler form by dividing the entire expression by $x$. This substitution leads to:
$\left(\frac{1}{\sqrt{1-v^2}} + e^v\right) x \frac{dv}{dx} = 1$
Separating the variables results in the integral:
$\left(\frac{1}{\sqrt{1-v^2}} + e^v\right) dv = \frac{1}{x} dx$
Integration yields $\sin^{-1}(v) + e^v = \ln|x| + C$. Using the initial condition $y(1) = 0$, where $v = 0$ at $x = 1$, allows for the determination of the constant: $\sin^{-1}(0) + e^0 = \ln(1) + C$, which simplifies to $1 = C$. With the general solution $\sin^{-1}\left(\frac{y}{x}\right) + e^{\frac{y}{x}} = \ln(x) + 1$, the coordinate point $(2\alpha, \alpha)$ satisfies:
$\sin^{-1}\left(\frac{\alpha}{2\alpha}\right) + e^{\frac{\alpha}{2\alpha}} = \ln(2\alpha) + 1$
Since $\sin^{-1}(\frac{1}{2}) = \frac{\pi}{6}$ and $e^{\frac{1}{2}} = \sqrt{e}$, the equation becomes $\frac{\pi}{6} + \sqrt{e} = \ln(2\alpha) + 1$. Isolating $\ln(2\alpha)$ results in $\ln(2\alpha) = \frac{\pi}{6} + \sqrt{e} - 1$. Exponentiating both sides yields $2\alpha = \exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$, leading to $\alpha = \frac{1}{2} \exp\left(\frac{\pi}{6} + \sqrt{e} - 1\right)$.

This differential equation is homogeneous, which suggests substituting $y = vx$ to simplify the expression into a separable form. By rearranging the original equation as $\frac{dy}{dx} = \frac{\sqrt{x^{2+}y^2}}{x} + \frac{y}{x}$, we can substitute $y = vx$ and $\frac{dy}{dx} = v + x \frac{dv}{dx}$. This transformation leads to the equation $v + x \frac{dv}{dx} = \sqrt{1+v^2} + v$, which simplifies significantly to $\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$.

Integrating both sides of this equation yields $\ln(v + \sqrt{1+v^2}) = \ln x + C$. Applying the initial condition where the curve passes through $x=1$ at $y=0$, we find $v = 0$, leading to $\ln(1) = \ln(1) + C$, which confirms that $C=0$. Exponentiating both sides results in the relationship $v + \sqrt{1+v^2} = x$. Substituting $v = \frac{y}{x}$ back into this equation gives $\frac{y}{x} + \sqrt{1 + \frac{y^2}{x^2}} = x$. To find the value of $\alpha$ when $x=2$, we substitute the coordinates $(2, \alpha)$ into this result to get $\frac{\alpha}{2} + \sqrt{1 + \frac{\alpha^2}{4}} = 2$. Multiplying the entire equation by two yields $\alpha + \sqrt{4 + \alpha^2} = 4$, which can be rearranged as $\sqrt{4 + \alpha^2} = 4 - \alpha$. Squaring both sides produces $4 + \alpha^2 = 16 - 8\alpha + \alpha^2$, which simplifies to $8\alpha = 12$, resulting in $\alpha = \frac{3}{2}$.

The differential equation $y^2 dx + (x^2 - xy + y^2) dy = 0$ is a homogeneous equation, which can be solved efficiently by substituting $y = vx$, where the derivative $\frac{dy}{dx}$ becomes $v + x \frac{dv}{dx}$. Substituting these into the original equation leads to the expression $v + x \frac{dv}{dx} = \frac{v^2}{v - 1 - v^2}$, which rearranges to the separable form $\int \frac{v - 1 - v^2}{v(1 + v^2)} dv = \int \frac{dx}{x}$. The integrand on the left simplifies to the form $\int (\frac{1}{1 + v^2} - \frac{1}{v}) dv$, making the integration straightforward and resulting in $\tan^{-1}(v) - \ln|v| = \ln|x| + c$. Replacing $v$ with $\frac{y}{x}$ gives the general solution $\tan^{-1}(\frac{y}{x}) - \ln(\frac{y}{x}) = \ln(x) + c$.

Applying the condition that the curve passes through the point (1, 1), we find that $\tan^{-1}(1) - \ln(1) = \ln(1) + c$, which determines the constant $c = \frac{\pi}{4}$. Given that the curve intersects the line $y = \sqrt{3}x$ at the point $(\alpha, \sqrt{3}\alpha)$, the ratio $\frac{y}{x}$ at this point is $\sqrt{3}$. Substituting these values into our equation yields $\tan^{-1}(\sqrt{3}) - \ln(\sqrt{3}) = \ln(\alpha) + \frac{\pi}{4}$. Since $\tan^{-1}(\sqrt{3})$ is $\frac{\pi}{3}$, the expression simplifies to $\frac{\pi}{3} - \ln(\sqrt{3}) = \ln(\alpha) + \frac{\pi}{4}$, which allows us to solve for $\ln(\alpha)$ as $\frac{\pi}{12} - \ln(\sqrt{3})$. Consequently, the value of $\log_e(\sqrt{3}\alpha)$, which is $\ln(\sqrt{3}) + \ln(\alpha)$, equals $\ln(\sqrt{3}) + \frac{\pi}{12} - \ln(\sqrt{3})$, leaving a final result of $\frac{\pi}{12}$.

The relationship between the area under a curve and its function value allows us to express the region as
$\int_{3}^{x} y(t) dt = \left(\frac{y}{x}\right)^3$
To transform this into a differential equation, we multiply both sides by $x^3$ to get $y^3 = x^3 \int_{3}^{x} y(t) dt$ and differentiate with respect to $x$. Applying the product rule to the right side yields $3y^2 \frac{dy}{dx} = 3x^2 \int_{3}^{x} y(t) dt + x^3 y$. Substituting the initial integral definition back into this derivative leads to $3y^2 \frac{dy}{dx} = 3x^2 \left(\frac{y^3}{x^3}\right) + x^3 y$, which simplifies to $3y^2 \frac{dy}{dx} = \frac{3y^3}{x} + x^3 y$.

After dividing the entire expression by $y$, we arrive at $3y \frac{dy}{dx} = \frac{3y^2}{x} + x^3$. We simplify this by substituting $v = y^2$, which implies $\frac{dv}{dx} = 2y \frac{dy}{dx}$. Rewriting the equation using this substitution transforms it into the linear form
$\frac{dv}{dx} - \frac{2}{x} v = \frac{2}{3} x^3$
Using an integrating factor of $e^{\int -\frac{2}{x} dx} = \frac{1}{x^2}$, we solve for $v$ by integrating both sides, resulting in $\frac{v}{x^2} = \int \frac{2}{3} x dx = \frac{x^2}{3} + C$. Converting back to $y$ gives $y^2 = \frac{x^4}{3} + Cx^2$.

Using the condition that the curve passes through $(3,3)$, we find $3^2 = \frac{3^4}{3} + 9C$, which leads to $9 = 27 + 9C$ and identifies $C = -2$. The curve is thus represented by $y^2 = \frac{x^4}{3} - 2x^2$. Evaluating this at the point $(\alpha, 6\sqrt{10})$ requires solving $(6\sqrt{10})^2 = \frac{\alpha^4}{3} - 2\alpha^2$, or $360 = \frac{\alpha^4}{3} - 2\alpha^2$. Multiplying by 3 gives $\alpha^4 - 6\alpha^2 - 1080 = 0$. Solving this quadratic equation for $\alpha^2$ yields 36, confirming that $\alpha = 6$ for the point in the first quadrant.

Integrating the differential equation $\frac{dy}{dx} = \frac{1}{1+\sin 2x}$ by rewriting the denominator as $(\sin x + \cos x)^2$ yields the general solution $y = -\frac{1}{1+\tan x} + c$. Applying the condition $y(\frac{\pi}{4}) = \frac{1}{2}$ determines the constant $c = 1$, resulting \in the simplified function $y = \frac{\sin x}{\sin x + \cos x}$. Setting this equal to the curve $y = \sqrt{2}\sin x$ results \in the intersection points $x = \pi$ or the solutions to $\sin(x+\frac{\pi}{4}) = \frac{1}{2}$, which are $x = \frac{7\pi}{12}$ and $x = \frac{23\pi}{12}$. Summing these abscissas gives $\pi + \frac{7\pi}{12} + \frac{23\pi}{12} = \frac{42\pi}{12}$, which identifies the value of $k$ as 42.