Linear Differential Equations

Introduction to Linear Differential Equations

A first-order differential equation is said to be linear if the dependent variable (usually $y$) and its derivative $\frac{dy}{dx}$ appear only in the first degree and are not multiplied together.

The standard form of a first-order linear differential equation is: $$\frac{dy}{dx} + P(x)\,y = Q(x)$$ Here, $P(x)$ and $Q(x)$ are constants or functions of $x$ only.

A few important observations:

• $\frac{dy}{dx} + 3y = e^x$ is linear.
• $\frac{dy}{dx} + y^2 = x$ is not linear because $y$ is squared.
• $y\frac{dy}{dx} = x$ is not linear because $y$ and $\frac{dy}{dx}$ are multiplied.

So in a linear differential equation, both $y$ and $\frac{dy}{dx}$ must occur only to the first power, and the coefficient of $y$ must depend only on $x$.

The Integrating Factor (I.F.)

To solve a linear differential equation, we multiply the whole equation by a special function called the Integrating Factor (I.F.).

For the equation $$\frac{dy}{dx} + P(x)y = Q(x),$$ the Integrating Factor is $$\text{I.F.} = e^{\int P(x)\,dx}.$$

Why is this useful? Because after multiplying the equation by the I.F., the left-hand side becomes the exact derivative of the product $y \cdot \text{I.F.}$: $$\text{I.F.}\cdot \frac{dy}{dx} + P(x)\cdot \text{I.F.}\cdot y = \frac{d}{dx}\big(y\cdot \text{I.F.}\big).$$

This is the key idea behind the method.

Important logarithm rule

When simplifying the I.F., expressions of the form $$e^{\ln f(x)}$$ reduce to $$f(x),$$ provided $f(x) > 0$; in applied textbook work we often write the simplified algebraic form directly. For example, $$e^{\int \frac{2}{x}dx} = e^{2\ln|x|} = e^{\ln(x^2)} = x^2.$$

The General Solution

Once the Integrating Factor is found, multiply the equation $$\frac{dy}{dx} + P(x)y = Q(x)$$ by the I.F. Then the equation becomes $$\frac{d}{dx}\big(y\cdot \text{I.F.}\big) = Q(x)\cdot \text{I.F.}.$$ Integrating both sides with respect to $x$, we get $$y\cdot (\text{I.F.}) = \int Q(x)\cdot (\text{I.F.})\,dx + C.$$

This is the standard formula for the general solution.

Step-by-step algorithm

1. Write the equation in the form $\frac{dy}{dx} + P(x)y = Q(x)$.
2. Identify $P(x)$ and $Q(x)$.
3. Find the Integrating Factor: $$\text{I.F.} = e^{\int P(x)dx}.$$
4. Use $$y\cdot (\text{I.F.}) = \int Q(x)\cdot (\text{I.F.})\,dx + C.$$
5. Simplify the result and, if needed, solve explicitly for $y$.

This method works only after the equation has been reduced to standard linear form.

The Alternate Form (Linear in $x$)

Sometimes the given differential equation is not linear in $y$, but it is linear in $x$ if we treat $x$ as the dependent variable and $y$ as the independent variable.

Then the standard form becomes $$\frac{dx}{dy} + P_1(y)x = Q_1(y).$$ Here, $P_1(y)$ and $Q_1(y)$ are constants or functions of $y$ only.

The method is exactly analogous:

• Integrating Factor: $$\text{I.F.} = e^{\int P_1(y)\,dy}$$
• General solution: $$x\cdot (\text{I.F.}) = \int Q_1(y)\cdot (\text{I.F.})\,dy + C$$

So the core idea remains the same; only the roles of $x$ and $y$ are interchanged.

Example 1: Standard Linear Equation

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

Solution: Step 1: Compare with the standard form $$\frac{dy}{dx} + P y = Q.$$ Here, $$P = 2, \qquad Q = e^{3x}.$$

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

Step 3: Multiply the whole equation by $e^{2x}$. $$e^{2x}\frac{dy}{dx} + 2e^{2x}y = e^{5x}.$$ The left-hand side is $$\frac{d}{dx}(ye^{2x}).$$ So, $$\frac{d}{dx}(ye^{2x}) = e^{5x}.$$

Step 4: Integrate both sides. $$ye^{2x} = \int e^{5x}\,dx + C = \frac{e^{5x}}{5} + C.$$

Step 5: Solve for $y$. $$y = \frac{e^{5x}}{5e^{2x}} + Ce^{-2x} = \frac{e^{3x}}{5} + Ce^{-2x}.$$

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

Example 2: Normalizing the Equation

Solve: $x \frac{dy}{dx} + 2y = x^2 \quad (x \neq 0)$.

Solution: Step 1: The coefficient of $\frac{dy}{dx}$ must be 1. Divide the entire equation by $x$: $$\frac{dy}{dx} + \frac{2}{x}y = x.$$ So, $$P = \frac{2}{x}, \qquad Q = x.$$

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

Step 3: Multiply the equation by $x^2$: $$x^2\frac{dy}{dx} + 2xy = x^3.$$ The left side is $$\frac{d}{dx}(x^2y).$$ Hence, $$\frac{d}{dx}(x^2y) = x^3.$$

Step 4: Integrate: $$x^2y = \int x^3\,dx + C = \frac{x^4}{4} + C.$$

Step 5: Explicit form, if required: $$y = \frac{x^2}{4} + \frac{C}{x^2}.$$

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

Example 3: Trigonometric Linear Equation

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

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

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

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

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

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

Example 4: Initial Value Problem (IVP)

Find the particular solution of $\frac{dy}{dx} + y \cot x = 2x + x^2 \cot x$, given that $y = 0$ when $x = \frac{\pi}{2}$.

Solution: Step 1: Identify $$P = \cot x, \qquad Q = 2x + x^2\cot x.$$

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

Step 3: Multiply through by $\sin x$: $$y\sin x = \int (2x + x^2\cot x)\sin x\,dx + C.$$ Since $\cot x \cdot \sin x = \cos x$, this becomes $$y\sin x = \int (2x\sin x + x^2\cos x)\,dx + C.$$

Step 4: Recognize the integrand as an exact derivative. $$\frac{d}{dx}(x^2\sin x) = 2x\sin x + x^2\cos x.$$ Therefore, $$\int (2x\sin x + x^2\cos x)\,dx = x^2\sin x.$$ So the general solution is $$y\sin x = x^2\sin x + C.$$

Step 5: Use the condition $y=0$ when $x=\frac{\pi}{2}$. $$0\cdot \sin\left(\frac{\pi}{2}\right) = \left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + C$$ $$0 = \frac{\pi^2}{4} + C$$ $$C = -\frac{\pi^2}{4}.$$

Step 6: Substitute back: $$y\sin x = x^2\sin x - \frac{\pi^2}{4}.$$

Answer: $y\sin x = x^2\sin x - \frac{\pi^2}{4}$

Example 5: Linear in $x$ (Alternate Form)

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

Solution: Step 1: Divide by $(y^2+1)$ to get standard linear form in $x$: $$\frac{dx}{dy} + \frac{1}{y^2+1}x = \frac{e^{\tan^{-1} y}}{y^2+1}.$$ Thus, $$P_1 = \frac{1}{y^2+1}, \qquad Q_1 = \frac{e^{\tan^{-1} y}}{y^2+1}.$$

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

Step 3: Apply the solution formula: $$x e^{\tan^{-1} y} = \int \frac{e^{\tan^{-1} y}}{y^2+1} \cdot e^{\tan^{-1} y}\,dy + C$$ $$x e^{\tan^{-1} y} = \int \frac{e^{2\tan^{-1} y}}{y^2+1}\,dy + C.$$

Step 4: Use substitution. Let $$t = \tan^{-1} y \implies dt = \frac{1}{y^2+1}dy.$$ Then the integral becomes $$\int e^{2t}\,dt = \frac{e^{2t}}{2} = \frac{e^{2\tan^{-1} y}}{2}.$$

Step 5: Final solution: $$x e^{\tan^{-1} y} = \frac{1}{2}e^{2\tan^{-1} y} + C.$$

Answer: $x e^{\tan^{-1} y} = \frac{1}{2} e^{2\tan^{-1} y} + C$

Example 6: Finding the Integrating Factor

Find the Integrating Factor of the differential equation: $x \ln x \frac{dy}{dx} + y = 2 \ln x$.

Solution: Step 1: Divide by $x\ln x$ to normalize the equation: $$\frac{dy}{dx} + \frac{1}{x\ln x}y = \frac{2}{x}.$$ So, $$P(x) = \frac{1}{x\ln x}.$$

Step 2: Compute 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 $$\int \frac{1}{x\ln x}dx = \int \frac{1}{t}dt = \ln|t| = \ln(\ln x).$$ Hence, $$\text{I.F.} = e^{\ln(\ln x)} = \ln x.$$

Answer: I.F. $= \ln x$

Example 7: Involving Inverse Trigonometry

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

Solution: Step 1: Divide by $(1-x^2)$: $$\frac{dy}{dx} - \frac{x}{1-x^2}y = \frac{1}{1-x^2}.$$ So, $$P = -\frac{x}{1-x^2}, \qquad Q = \frac{1}{1-x^2}.$$

Step 2: Find the Integrating Factor. $$\int P\,dx = \int -\frac{x}{1-x^2}\,dx.$$ Let $$t = 1-x^2 \implies dt = -2x\,dx.$$ Then $$\int -\frac{x}{1-x^2}\,dx = \frac{1}{2}\int \frac{dt}{t} = \frac{1}{2}\ln|t| = \frac{1}{2}\ln|1-x^2|.$$ Hence, $$\text{I.F.} = e^{\frac{1}{2}\ln|1-x^2|} = \sqrt{|1-x^2|}.$$ In the standard textbook interval where $1-x^2>0$, this is written as $$\text{I.F.} = \sqrt{1-x^2}.$$

Step 3: Apply the formula: $$y\sqrt{1-x^2} = \int \frac{1}{1-x^2}\sqrt{1-x^2}\,dx + C$$ $$y\sqrt{1-x^2} = \int \frac{1}{\sqrt{1-x^2}}\,dx + C.$$

Step 4: Integrate: $$\int \frac{1}{\sqrt{1-x^2}}\,dx = \sin^{-1}x.$$ So, $$y\sqrt{1-x^2} = \sin^{-1}x + C.$$

Answer: $y\sqrt{1-x^2} = \sin^{-1}x + C$

Example 8: A Tricky Substitution Leading to an LDE

Solve: $\frac{dy}{dx} + y = y^2 e^x$. (This is a Bernoulli differential equation, reducible to a Linear Differential Equation.)

Solution: Step 1: Divide the equation by $y^2$: $$y^{-2}\frac{dy}{dx} + y^{-1} = e^x.$$

Step 2: Use the substitution $$v = y^{-1}.$$ Then $$\frac{dv}{dx} = -y^{-2}\frac{dy}{dx} \implies y^{-2}\frac{dy}{dx} = -\frac{dv}{dx}.$$ Substitute into the equation: $$-\frac{dv}{dx} + v = e^x.$$ Multiply by $-1$: $$\frac{dv}{dx} - v = -e^x.$$ This is linear in $v$.

Step 3: Identify $$P = -1, \qquad Q = -e^x.$$ So, $$\text{I.F.} = e^{\int -1\,dx} = e^{-x}.$$

Step 4: Apply the formula: $$v e^{-x} = \int (-e^x)(e^{-x})\,dx + C = \int (-1)\,dx + C$$ $$v e^{-x} = -x + C.$$

Step 5: Replace $v = 1/y$: $$\frac{e^{-x}}{y} = C - x.$$ Multiply by $y$: $$e^{-x} = y(C-x).$$

Answer: $y(C - x) = e^{-x}$

Example 9: Complex Integration

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

Solution: Step 1: Divide by $x$: $$\frac{dy}{dx} + \frac{1}{x}y = \cos x + \frac{\sin x}{x}.$$ Thus, $$P = \frac{1}{x}, \qquad Q = \cos x + \frac{\sin x}{x}.$$

Step 2: Find the Integrating Factor: $$\text{I.F.} = e^{\int \frac{1}{x}\,dx} = e^{\ln|x|} = x.$$

Step 3: Apply the formula: $$xy = \int \left(\cos x + \frac{\sin x}{x}\right)x\,dx + C$$ $$xy = \int (x\cos x + \sin x)\,dx + C.$$

Step 4: Recognize the integrand: $$\frac{d}{dx}(x\sin x) = x\cos x + \sin x.$$ So, $$xy = x\sin x + C.$$

Answer: $xy = x\sin x + C$

Example 10: Recognizing Alternate Form

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

Solution: Step 1: Solving directly for $\frac{dy}{dx}$ does not give a linear equation in $y$. So invert the derivative: $$\frac{dx}{dy} = \frac{x+2y^3}{y} = \frac{x}{y} + 2y^2.$$

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

Step 3: Find the Integrating Factor with respect to $y$: $$\text{I.F.} = e^{\int -\frac{1}{y}\,dy} = e^{-\ln|y|} = \frac{1}{|y|}.$$ In the standard working form, this is written as $$\text{I.F.} = \frac{1}{y}.$$

Step 4: Apply the formula: $$x\cdot \frac{1}{y} = \int 2y^2\cdot \frac{1}{y}\,dy + C$$ $$\frac{x}{y} = \int 2y\,dy + C = y^2 + C.$$

Step 5: Multiply by $y$: $$x = y^3 + Cy.$$

Answer: $x = y^3 + Cy$