Solved Examples

Important Theorems and Concepts Recap

1. Order and Degree

• Order: The order of the highest derivative appearing in the differential equation.
• Degree: The power of the highest order derivative, provided the differential equation is a polynomial equation in terms of its derivatives.

2. Types of Solutions

• General Solution: Contains arbitrary constants equal in number to the order of the differential equation. Represents a family of curves.
• Particular Solution: Obtained by giving specific values to the arbitrary constants based on initial conditions.

3. Methods of Solving First-Order, First-Degree ODEs

• Variable Separable: $f(y) \, dy = g(x) \, dx \implies \int f(y) \, dy = \int g(x) \, dx + C$.
• Homogeneous: $\frac{dy}{dx} = F(x,y)$ where $F(\lambda x, \lambda y) = \lambda^0 F(x,y)$. Use substitution $y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$.
• Linear Differential Equation (LDE): $\frac{dy}{dx} + P(x)y = Q(x)$. Integrating Factor (I.F.) $= e^{\int P(x) \, dx}$. Solution: $y \cdot \text{(I.F.)} = \int Q(x) \cdot \text{(I.F.)} \, dx + C$.

These are the key tools repeatedly used in the solved examples below.

Example 1: Order and Degree

Find the order and degree of the differential equation: $\frac{d^2y}{dx^2} + 5\left(\frac{dy}{dx}\right)^3 + 6y = 0$.

Solution: Step 1: Identify the highest order derivative in the equation. The derivatives present are $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$. The highest order derivative is $\frac{d^2y}{dx^2}$. Therefore, the order is 2.

Step 2: Check whether the equation is polynomial in derivatives. Yes. All derivatives appear with non-negative integer powers, and no derivative is inside a trigonometric, logarithmic, or exponential function.

Step 3: Find the power of the highest order derivative. The term $\frac{d^2y}{dx^2}$ occurs to the first power. Therefore, the degree is 1.

Answer: Order = 2, Degree = 1.

Example 2: Order and Degree (Fractional Power)

Find the order and degree of: $\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right]^{3/2} = k \frac{d^2y}{dx^2}$.

Solution: Step 1: Identify the highest order derivative. The highest order derivative present is $\frac{d^2y}{dx^2}$. Hence, the order is 2.

Step 2: To determine the degree, first remove the fractional power so that the equation becomes polynomial in derivatives. Squaring both sides gives: $$\left[ 1 + \left(\frac{dy}{dx}\right)^2 \right]^3 = k^2 \left(\frac{d^2y}{dx^2}\right)^2$$

Step 3: Now the equation is polynomial in derivatives. The highest order derivative is still $\frac{d^2y}{dx^2}$, and its power is now 2. Therefore, the degree is 2.

Answer: Order = 2, Degree = 2.

Example 3: Order and Degree (Undefined)

Find the order and degree of: $\frac{d^3y}{dx^3} + y^2 + e^{dy/dx} = 0$.

Solution: Step 1: Identify the highest order derivative. The highest derivative is $\frac{d^3y}{dx^3}$. Therefore, the order is 3.

Step 2: Check whether the equation is polynomial in derivatives. It is not polynomial in derivatives because the first derivative $\frac{dy}{dx}$ appears inside the exponential function $e^{dy/dx}$.

Step 3: Since the equation is not polynomial in derivatives, the degree cannot be defined.

Answer: Order = 3, Degree = Not defined.

Example 4: Verifying a Solution

Verify that $y = A\cos x + B\sin x$ is a solution of $\frac{d^2y}{dx^2} + y = 0$.

Solution: Step 1: Differentiate $y$ with respect to $x$. $$y' = -A\sin x + B\cos x$$

Step 2: Differentiate again. $$y'' = -A\cos x - B\sin x$$

Step 3: Substitute into the left-hand side of the differential equation. $$y'' + y = (-A\cos x - B\sin x) + (A\cos x + B\sin x)$$ $$= 0$$

Step 4: Since the left-hand side becomes zero identically, the given function satisfies the equation.

Answer: Verified. Hence $y = A\cos x + B\sin x$ is a solution of $\frac{d^2y}{dx^2} + y = 0$.

Example 5: Basic Variable Separable

Solve: $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$.

Solution: Step 1: Separate the variables. $$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$$

Step 2: Integrate both sides. $$\int \frac{1}{1+y^2} \, dy = \int \frac{1}{1+x^2} \, dx$$

Step 3: Use the standard formula $\int \frac{1}{1+t^2}dt = \tan^{-1} t$. $$\tan^{-1} y = \tan^{-1} x + C$$

Answer: $\tan^{-1} y = \tan^{-1} x + C$

Example 6: Variable Separable with Exponentials

Solve: $\frac{dy}{dx} = e^{x-y} + x^2 e^{-y}$.

Solution: Step 1: Factor out $e^{-y}$ from the right-hand side. $$\frac{dy}{dx} = e^{-y}(e^x + x^2)$$

Step 2: Separate the variables. Multiply both sides by $e^y dx$: $$e^y \, dy = (e^x + x^2) \, dx$$

Step 3: Integrate both sides. $$\int e^y \, dy = \int (e^x + x^2) \, dx$$

Step 4: Evaluate the integrals. $$e^y = e^x + \frac{x^3}{3} + C$$

Answer: $e^y = e^x + \frac{x^3}{3} + C$

Example 7: Variable Separable Initial Value Problem

Solve: $\frac{dy}{dx} = -4xy^2$, given $y(0) = 1$.

Solution: Step 1: Separate variables. $$\frac{dy}{y^2} = -4x \, dx$$ $$y^{-2} \, dy = -4x \, dx$$

Step 2: Integrate both sides. $$\int y^{-2} \, dy = -4 \int x \, dx$$ $$-\frac{1}{y} = -2x^2 + C$$

Step 3: Apply the initial condition $y(0)=1$. Substitute $x=0$, $y=1$: $$-1 = 0 + C \implies C = -1$$

Step 4: Substitute the constant back. $$-\frac{1}{y} = -2x^2 - 1$$ Multiply by $-1$: $$\frac{1}{y} = 2x^2 + 1$$ Thus, $$y = \frac{1}{2x^2 + 1}$$

Answer: $y = \frac{1}{2x^2 + 1}$

Example 8: Equation Reducible to Variable Separable

Solve: $\frac{dy}{dx} = (x+y)^2$.

Solution: Step 1: Use the substitution $$v = x+y$$ Then, $$\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1$$

Step 2: Substitute into the differential equation. $$\frac{dv}{dx} - 1 = v^2$$ $$\frac{dv}{dx} = v^2 + 1$$

Step 3: Separate variables. $$\frac{dv}{1+v^2} = dx$$

Step 4: Integrate. $$\int \frac{dv}{1+v^2} = \int dx$$ $$\tan^{-1} v = x + C$$

Step 5: Replace $v$ by $x+y$. $$\tan^{-1}(x+y) = x + C$$

Answer: $\tan^{-1}(x+y) = x + C$

Example 9: Basic Homogeneous Equation

Solve: $(x+y)dy = (x-y)dx$.

Solution: Step 1: Rewrite in differential equation form. $$\frac{dy}{dx} = \frac{x-y}{x+y}$$ Since numerator and denominator are both homogeneous expressions of degree 1, the equation is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = \frac{x-vx}{x+vx} = \frac{1-v}{1+v}$$

Step 3: Simplify. $$x\frac{dv}{dx} = \frac{1-v}{1+v} - v = \frac{1-v-v-v^2}{1+v} = \frac{1-2v-v^2}{1+v}$$ So, $$\frac{1+v}{1-2v-v^2} \, dv = \frac{dx}{x}$$ Equivalently, $$\frac{v+1}{v^2+2v-1} \, dv = -\frac{dx}{x}$$

Step 4: Integrate. Let $$u = v^2 + 2v - 1 \implies du = 2(v+1) \, dv$$ Therefore, $$\frac{1}{2} \int \frac{du}{u} = -\int \frac{dx}{x}$$ $$\frac{1}{2}\ln|v^2+2v-1| = -\ln|x| + \ln|C|$$

Step 5: Simplify. $$\ln|v^2+2v-1| = -2\ln|x| + \ln|K|$$ $$v^2+2v-1 = \frac{K}{x^2}$$

Step 6: Substitute back $v = y/x$. $$\frac{y^2}{x^2} + \frac{2y}{x} - 1 = \frac{K}{x^2}$$ Multiply by $x^2$: $$y^2 + 2xy - x^2 = K$$

Answer: $y^2 + 2xy - x^2 = K$

Example 10: Homogeneous Equation with Radicals

Solve: $x \frac{dy}{dx} = y + \sqrt{x^2+y^2}$.

Solution: Step 1: Divide by $x$: $$\frac{dy}{dx} = \frac{y}{x} + \frac{\sqrt{x^2+y^2}}{x} = \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2}$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = v + \sqrt{1+v^2}$$

Step 3: Separate variables. $$x\frac{dv}{dx} = \sqrt{1+v^2}$$ $$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$$

Step 4: Integrate. $$\int \frac{dv}{\sqrt{1+v^2}} = \int \frac{dx}{x}$$ $$\ln|v+\sqrt{1+v^2}| = \ln|x| + \ln|C|$$ $$v+\sqrt{1+v^2} = Cx$$

Step 5: Replace $v = y/x$. $$\frac{y}{x} + \sqrt{1+\frac{y^2}{x^2}} = Cx$$ Multiply through by $x$: $$y + \sqrt{x^2+y^2} = Cx^2$$

Answer: $y + \sqrt{x^2+y^2} = Cx^2$

Example 11: Homogeneous using $x = vy$

Solve: $2y e^{x/y} dx + (y - 2x e^{x/y}) dy = 0$.

Solution: Step 1: Rewrite in the form $\frac{dx}{dy}$: $$2y e^{x/y} dx = (2x e^{x/y} - y) dy$$ $$\frac{dx}{dy} = \frac{2x e^{x/y} - y}{2y e^{x/y}}$$ Because the equation involves $e^{x/y}$, it is convenient to use $$x = vy$$ so that $$\frac{dx}{dy} = v + y\frac{dv}{dy}$$

Step 2: Substitute. $$v + y\frac{dv}{dy} = \frac{2vye^v - y}{2ye^v} = \frac{2ve^v - 1}{2e^v}$$

Step 3: Simplify. $$y\frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} - v = -\frac{1}{2e^v}$$

Step 4: Separate variables. $$2e^v \, dv = -\frac{dy}{y}$$

Step 5: Integrate. $$\int 2e^v \, dv = -\int \frac{dy}{y}$$ $$2e^v = -\ln|y| + C$$

Step 6: Replace $v = x/y$. $$2e^{x/y} + \ln|y| = C$$

Answer: $2e^{x/y} + \ln|y| = C$

Example 12: Basic Linear Differential Equation

Solve: $\frac{dy}{dx} + 2y = e^{3x}$.

Solution: Step 1: Compare with $$\frac{dy}{dx} + Py = Q$$ So, $$P = 2, \qquad Q = e^{3x}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int 2 \, dx} = e^{2x}$$

Step 3: Apply the linear DE formula. $$y e^{2x} = \int e^{3x} e^{2x} \, dx + C = \int e^{5x} \, dx + C$$

Step 4: Integrate. $$y e^{2x} = \frac{e^{5x}}{5} + C$$ Hence, $$y = \frac{e^{3x}}{5} + Ce^{-2x}$$

Answer: $y = \frac{e^{3x}}{5} + Ce^{-2x}$

Example 13: Linear DE Requiring Normalization

Solve: $x \frac{dy}{dx} + 2y = x^2$.

Solution: Step 1: Divide by $x$ to get standard form. $$\frac{dy}{dx} + \frac{2}{x}y = x$$ Thus, $$P = \frac{2}{x}, \qquad Q = x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{2}{x} \, dx} = e^{2\ln|x|} = x^2$$

Step 3: Use the formula. $$yx^2 = \int x\cdot x^2 \, dx + C = \int x^3 \, dx + C$$

Step 4: Integrate. $$yx^2 = \frac{x^4}{4} + C$$

Answer: $yx^2 = \frac{x^4}{4} + C$

Example 14: Trigonometric Linear Equation

Solve: $\frac{dy}{dx} + y\sec x = \tan x$.

Solution: Step 1: Here, $$P = \sec x, \qquad Q = \tan x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \sec x \, dx} = e^{\ln|\sec x + \tan x|} = \sec x + \tan x$$

Step 3: Apply the formula. $$y(\sec x + \tan x) = \int \tan x(\sec x + \tan x) \, dx + C$$ $$= \int (\sec x\tan x + \tan^2 x) \, dx + C$$ Use $$\tan^2 x = \sec^2 x - 1$$ So, $$\int (\sec x\tan x + \sec^2 x - 1) \, dx$$

Step 4: Integrate termwise. $$\int \sec x\tan x \, dx = \sec x$$ $$\int \sec^2 x \, dx = \tan x$$ $$\int 1 \, dx = x$$ Therefore, $$y(\sec x + \tan x) = \sec x + \tan x - x + C$$

Answer: $y(\sec x + \tan x) = \sec x + \tan x - x + C$

Example 15: Linear DE in $x$ (Alternate Form)

Solve: $(x + y)dy = dx$.

Solution: Step 1: Rewrite as $$\frac{dx}{dy} = x + y$$ This is linear in $x$: $$\frac{dx}{dy} - x = y$$ So, $$P_1 = -1, \qquad Q_1 = y$$

Step 2: Find the Integrating Factor with respect to $y$. $$\text{I.F.} = e^{\int -1 \, dy} = e^{-y}$$

Step 3: Apply the formula. $$x e^{-y} = \int y e^{-y} \, dy + C$$

Step 4: Integrate by parts. Take $$u = y, \quad dv = e^{-y}dy$$ Then $$du = dy, \quad v = -e^{-y}$$ So, $$\int y e^{-y}dy = -ye^{-y} + \int e^{-y}dy = -ye^{-y} - e^{-y} + C$$ Thus, $$x e^{-y} = -ye^{-y} - e^{-y} + C$$

Step 5: Multiply by $e^y$. $$x = -y - 1 + Ce^y$$ Rearranging, $$x + y + 1 = Ce^y$$

Answer: $x + y + 1 = Ce^y$

Example 16: Linear IVP

Solve: $\cos^2 x \frac{dy}{dx} + y = \tan x$, given $y\left(\frac{\pi}{4}\right) = 0$.

Solution: Step 1: Divide by $\cos^2 x$: $$\frac{dy}{dx} + y\sec^2 x = \tan x\sec^2 x$$ Thus, $$P = \sec^2 x, \qquad Q = \tan x\sec^2 x$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \sec^2 x \, dx} = e^{\tan x}$$

Step 3: Use the formula. $$y e^{\tan x} = \int \tan x\sec^2 x \, e^{\tan x} dx + C$$ Let $$t = \tan x \implies dt = \sec^2 x \, dx$$ Then the integral becomes $$\int t e^t \, dt$$ By integration by parts, $$\int t e^t \, dt = te^t - e^t$$ Therefore, $$y e^{\tan x} = e^{\tan x}(\tan x - 1) + C$$

Step 4: Apply the initial condition $y(\pi/4)=0$. At $x=\pi/4$, we have $\tan(\pi/4)=1$. So, $$0 = e^1(1-1) + C \implies C = 0$$

Step 5: Hence, $$y e^{\tan x} = e^{\tan x}(\tan x - 1)$$ $$y = \tan x - 1$$

Answer: $y = \tan x - 1$

JEE Main Standard

Example 17: Bernoulli's Equation

Solve: $x \frac{dy}{dx} + y = x^3 y^6$.

Solution: Step 1: Divide by $x$: $$\frac{dy}{dx} + \frac{1}{x}y = x^2 y^6$$ This is a Bernoulli equation of the form $$\frac{dy}{dx} + Py = Qy^n, \quad n=6$$

Step 2: Divide by $y^6$: $$y^{-6}\frac{dy}{dx} + \frac{1}{x}y^{-5} = x^2$$ Now put $$v = y^{-5}$$ Then $$\frac{dv}{dx} = -5y^{-6}\frac{dy}{dx} \implies y^{-6}\frac{dy}{dx} = -\frac{1}{5}\frac{dv}{dx}$$

Step 3: Substitute. $$-\frac{1}{5}\frac{dv}{dx} + \frac{1}{x}v = x^2$$ Multiply by $-5$: $$\frac{dv}{dx} - \frac{5}{x}v = -5x^2$$ This is linear in $v$.

Step 4: Find the Integrating Factor. $$\text{I.F.} = e^{\int -\frac{5}{x} \, dx} = e^{-5\ln|x|} = x^{-5}$$

Step 5: Apply the formula. $$v x^{-5} = \int (-5x^2)(x^{-5}) \, dx + C = \int -5x^{-3} \, dx + C$$ $$v x^{-5} = \frac{5}{2}x^{-2} + C$$

Step 6: Replace $v = y^{-5}$. $$y^{-5}x^{-5} = \frac{5}{2x^2} + C$$ That is, $$\frac{1}{x^5 y^5} = \frac{5}{2x^2} + C$$

Answer: $\frac{1}{x^5 y^5} = \frac{5}{2x^2} + C$

Example 18: Integration Factor Trick

Solve: $(x^2+1)\frac{dy}{dx} + 2xy = \sqrt{x^2+4}$.

Solution: Step 1: Observe the left-hand side carefully. Using the product rule, $$\frac{d}{dx}\big((x^2+1)y\big) = (x^2+1)\frac{dy}{dx} + 2xy$$ So the equation becomes $$\frac{d}{dx}\big((x^2+1)y\big) = \sqrt{x^2+4}$$

Step 2: Integrate both sides. $$(x^2+1)y = \int \sqrt{x^2+4} \, dx + C$$

Step 3: Use the standard integral formula $$\int \sqrt{x^2+a^2} \, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2}\ln|x+\sqrt{x^2+a^2}| + C$$ With $a=2$, we get $$(x^2+1)y = \frac{x}{2}\sqrt{x^2+4} + 2\ln|x+\sqrt{x^2+4}| + C$$

Answer: $(x^2+1)y = \frac{x}{2}\sqrt{x^2+4} + 2\ln|x+\sqrt{x^2+4}| + C$

Example 19: Variable Separable with Modulus

Solve: $\frac{dy}{dx} = |x|$, given $y(0) = 1$.

Solution: Step 1: Write in separable form. $$dy = |x| \, dx$$

Step 2: Integrate. $$y = \int |x| \, dx + C$$ We know $$\int |x| \, dx = \frac{x|x|}{2}$$ Hence, $$y = \frac{x|x|}{2} + C$$

Step 3: Apply the condition $y(0)=1$. $$1 = 0 + C \implies C = 1$$

Step 4: Therefore, $$y = \frac{x|x|}{2} + 1$$

Answer: $y = \frac{x|x|}{2} + 1$

Example 20: Homogeneous Form

Solve: $x \, dy - y \, dx = \sqrt{x^2+y^2} \, dx$.

Solution: Step 1: Rearrange to get $\frac{dy}{dx}$. $$x \, dy = (y + \sqrt{x^2+y^2}) \, dx$$ $$\frac{dy}{dx} = \frac{y + \sqrt{x^2+y^2}}{x} = \frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2}$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Then, $$v + x\frac{dv}{dx} = v + \sqrt{1+v^2}$$

Step 3: Separate variables. $$x\frac{dv}{dx} = \sqrt{1+v^2}$$ $$\frac{dv}{\sqrt{1+v^2}} = \frac{dx}{x}$$

Step 4: Integrate. $$\ln|v+\sqrt{1+v^2}| = \ln|x| + \ln|C|$$ $$v+\sqrt{1+v^2} = Cx$$

Step 5: Replace $v = y/x$. $$\frac{y}{x} + \sqrt{1+\left(\frac{y}{x}\right)^2} = Cx$$ Multiply by $x$: $$y + \sqrt{x^2+y^2} = Cx^2$$

Answer: $y + \sqrt{x^2+y^2} = Cx^2$

Example 21: Substitution with $x+y$

Solve: $\frac{dy}{dx} = \sin(x+y) + \cos(x+y)$.

Solution: Step 1: Use the substitution $$v = x+y$$ Then, $$\frac{dv}{dx} = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = \frac{dv}{dx} - 1$$

Step 2: Substitute. $$\frac{dv}{dx} - 1 = \sin v + \cos v$$ $$\frac{dv}{dx} = 1 + \sin v + \cos v$$

Step 3: Separate variables. $$\frac{dv}{1+\sin v + \cos v} = dx$$

Step 4: Simplify the denominator using the half-angle substitution. Let $$t = \tan\frac{v}{2}$$ Then, $$\sin v = \frac{2t}{1+t^2}, \qquad \cos v = \frac{1-t^2}{1+t^2}, \qquad dv = \frac{2dt}{1+t^2}$$ Substitute into the integral: $$\int \frac{\frac{2dt}{1+t^2}}{1 + \frac{2t}{1+t^2} + \frac{1-t^2}{1+t^2}} = \int dx$$ The denominator simplifies to $$\frac{2+2t}{1+t^2}$$ So the left side becomes $$\int \frac{2dt}{2+2t} = \int \frac{dt}{1+t}$$ Thus, $$\ln|1+t| = x + C$$

Step 5: Replace $t$ and then $v$. $$\ln\left|1+\tan\frac{v}{2}\right| = x + C$$ $$\ln\left|1+\tan\frac{x+y}{2}\right| = x + C$$

Answer: $\ln\left|1+\tan\left(\frac{x+y}{2}\right)\right| = x + C$

Example 22: Linear DE with Logarithms

Solve: $x \ln x \frac{dy}{dx} + y = \frac{2}{x}$.

Solution: Step 1: Divide through by $x\ln x$. $$\frac{dy}{dx} + \frac{1}{x\ln x}y = \frac{2}{x^2 \ln x}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{1}{x\ln x} \, dx}$$ Let $$t = \ln x \implies dt = \frac{1}{x}dx$$ Then, $$\int \frac{1}{x\ln x}dx = \int \frac{1}{t}dt = \ln(\ln x)$$ Hence, $$\text{I.F.} = e^{\ln(\ln x)} = \ln x$$

Step 3: Apply the formula. $$y\ln x = \int \frac{2}{x^2\ln x}(\ln x) \, dx + C = \int \frac{2}{x^2} \, dx + C$$

Step 4: Integrate. $$y\ln x = -\frac{2}{x} + C$$ Rearranging, $$y\ln x + \frac{2}{x} = C$$

Answer: $y\ln x + \frac{2}{x} = C$

Example 23: Differential Equation from Given Slope

Find the curve passing through $(1, 1)$ given that the slope of the tangent at any point $(x, y)$ is $\frac{y}{x} - \cos^2\left(\frac{y}{x}\right)$.

Solution: Step 1: Write the differential equation. $$\frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right)$$ This is homogeneous.

Step 2: Put $$y = vx \implies \frac{dy}{dx} = v + x\frac{dv}{dx}$$ Substitute: $$v + x\frac{dv}{dx} = v - \cos^2 v$$

Step 3: Simplify. $$x\frac{dv}{dx} = -\cos^2 v$$ $$\sec^2 v \, dv = -\frac{dx}{x}$$

Step 4: Integrate. $$\tan v = -\ln|x| + C$$ Replace $v = y/x$: $$\tan\left(\frac{y}{x}\right) = -\ln|x| + C$$

Step 5: Apply the condition that the curve passes through $(1,1)$. $$\tan\left(\frac{1}{1}\right) = -\ln 1 + C$$ $$\tan 1 = C$$

Step 6: Therefore, $$\tan\left(\frac{y}{x}\right) + \ln|x| = \tan 1$$

Answer: $\tan\left(\frac{y}{x}\right) + \ln|x| = \tan 1$

Example 24: Application in Geometry

The y-intercept of the tangent at any point $P(x,y)$ on a curve is proportional to the square of the y-coordinate of $P$. Form the differential equation.

Solution: Step 1: Equation of tangent at $(x,y)$ is $$Y - y = \frac{dy}{dx}(X - x)$$

Step 2: To find the y-intercept, put $X=0$. Then, $$Y - y = -x\frac{dy}{dx}$$ So the y-intercept is $$Y = y - x\frac{dy}{dx}$$

Step 3: The problem states that this y-intercept is proportional to $y^2$. So, $$y - x\frac{dy}{dx} = ky^2$$ for some constant of proportionality $k$.

Step 4: Rearrange if desired. $$x\frac{dy}{dx} - y + ky^2 = 0$$

Answer: $x\frac{dy}{dx} - y + ky^2 = 0$

Example 25: Clever Transformation

Solve: $\frac{dy}{dx} = \frac{y^3}{e^{2x} + y^2}$.

Solution: Step 1: Rewrite the equation in a form suitable for substitution. Take reciprocal carefully by writing $$\frac{dx}{dy} = \frac{e^{2x} + y^2}{y^3} = \frac{e^{2x}}{y^3} + \frac{1}{y}$$ This is not linear in $x$, but a substitution involving $e^{-2x}$ will help.

Step 2: Rearrange: $$\frac{dx}{dy} - \frac{1}{y} = \frac{e^{2x}}{y^3}$$ Multiply both sides by $e^{-2x}$: $$e^{-2x}\frac{dx}{dy} - \frac{1}{y}e^{-2x} = y^{-3}$$

Step 3: Put $$v = e^{-2x}$$ Then, $$\frac{dv}{dy} = -2e^{-2x}\frac{dx}{dy} = -2v\frac{dx}{dy}$$ So, $$e^{-2x}\frac{dx}{dy} = -\frac{1}{2}\frac{dv}{dy}$$

Step 4: Substitute into the transformed equation. $$-\frac{1}{2}\frac{dv}{dy} - \frac{1}{y}v = y^{-3}$$ Multiply by $-2$: $$\frac{dv}{dy} + \frac{2}{y}v = -2y^{-3}$$ This is linear in $v$.

Step 5: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{2}{y} \, dy} = y^2$$

Step 6: Apply the formula. $$vy^2 = \int (-2y^{-3})y^2 \, dy + C = \int -2y^{-1} \, dy + C$$ $$vy^2 = -2\ln|y| + C$$

Step 7: Replace $v = e^{-2x}$. $$y^2 e^{-2x} = C - 2\ln|y|$$ Rearranging, $$y^2 e^{-2x} + 2\ln|y| = C$$

Answer: $y^2 e^{-2x} + 2\ln|y| = C$

Example 26: Separable with Advanced Integration

Solve: $\frac{dy}{dx} = \frac{\sin x + x\cos x}{y\ln y}$.

Solution: Step 1: Separate variables. $$y\ln y \, dy = (\sin x + x\cos x) \, dx$$

Step 2: Integrate the left side by parts. Let $$u = \ln y, \qquad dv = y \, dy$$ Then, $$du = \frac{1}{y}dy, \qquad v = \frac{y^2}{2}$$ So, $$\int y\ln y \, dy = \frac{y^2}{2}\ln y - \int \frac{y^2}{2}\cdot \frac{1}{y}dy$$ $$= \frac{y^2}{2}\ln y - \frac{1}{2}\int y \, dy$$ $$= \frac{y^2}{2}\ln y - \frac{y^2}{4}$$

Step 3: Integrate the right side. Notice that $$\frac{d}{dx}(x\sin x) = \sin x + x\cos x$$ Hence, $$\int (\sin x + x\cos x)dx = x\sin x + C$$

Step 4: Combine the results. $$\frac{y^2}{2}\ln y - \frac{y^2}{4} = x\sin x + C$$

Answer: $\frac{y^2}{2}\ln y - \frac{y^2}{4} = x\sin x + C$

Example 27: Homogeneous with Specific Boundary

Solve $x \, dy - y \, dx = \sqrt{x^2+y^2} \, dx$, given $y(1) = 0$.

Solution: Step 1: From Example 20, the general solution is $$y + \sqrt{x^2+y^2} = Cx^2$$

Step 2: Apply the boundary condition $y(1)=0$. Substitute $x=1$, $y=0$: $$0 + \sqrt{1^2+0^2} = C(1)^2$$ $$1 = C$$

Step 3: Therefore, the particular solution is $$y + \sqrt{x^2+y^2} = x^2$$

Answer: $y + \sqrt{x^2+y^2} = x^2$

Example 28: Form $\frac{dx}{dy} + Px = Q$

Solve: $(1+y^2)dx = (\tan^{-1}y - x)dy$.

Solution: Step 1: Rearrange into linear form in $x$: $$\frac{dx}{dy} = \frac{\tan^{-1}y - x}{1+y^2}$$ $$\frac{dx}{dy} + \frac{1}{1+y^2}x = \frac{\tan^{-1}y}{1+y^2}$$ Thus, $$P_1 = \frac{1}{1+y^2}, \qquad Q_1 = \frac{\tan^{-1}y}{1+y^2}$$

Step 2: Find the Integrating Factor. $$\text{I.F.} = e^{\int \frac{1}{1+y^2} \, dy} = e^{\tan^{-1}y}$$

Step 3: Apply the linear formula. $$x e^{\tan^{-1}y} = \int \frac{\tan^{-1}y}{1+y^2} e^{\tan^{-1}y}dy + C$$ Let $$t = \tan^{-1}y \implies dt = \frac{1}{1+y^2}dy$$ Then the integral becomes $$\int t e^t \, dt$$ By integration by parts, $$\int t e^t \, dt = te^t - e^t$$ So, $$x e^{\tan^{-1}y} = e^{\tan^{-1}y}(\tan^{-1}y - 1) + C$$

Step 4: Divide by $e^{\tan^{-1}y}$. $$x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$$

Answer: $x = \tan^{-1}y - 1 + Ce^{-\tan^{-1}y}$