# Linear Differential Equation

## Linear Differential Equation

Definition: An equation of the form dy/dx + Py = Q, where P and Q are function of x only, is called a Linear differential equation.

Solution of Linear Equation: Given equation is dy/dx + P y = Q

$${{e}^{\int{p.dx}}}\left( \frac{dy}{dx}+Py \right)=Q.{{e}^{\int{p.dx}}}$$,

$$\frac{d}{dx}y.{{e}^{\int{p.dx}}}=Q.{{e}^{\int{p.dx}}}$$,

$$\frac{d}{dx}y.{{e}^{\int{p.dx}}}.dx=Q.{{e}^{\int{p.dx}}}.dx$$,

$$y.{{e}^{\int{p.dx}}}=\int{Q.{{e}^{\int{p.dx}}}}dx+C$$.

Example 1: Solve (1 + x²) dy/dx + y = etan¯¹x

Solution: Given that (1 + x²) dy/dx + y = etan¯¹x

$$\frac{dy}{dx}+\frac{1}{1+{{x}^{2}}}.y=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$$,

$$P=\frac{1}{1+{{x}^{2}}}$$, $$Q=\frac{{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}}$$,

Integration factor (IF) = $${{e}^{\int{\frac{1}{1+{{x}^{2}}}dx}}}={{e}^{{{\tan }^{-1}}x}}$$,

Solution is $$y{{e}^{{{\tan }^{-1}}x}}=\int{\frac{{{e}^{2{{\tan }^{-1}}x}}}{1+{{x}^{2}}}dx}$$,

Put tan¯¹x = t

Then 1/ (1 + x²) dx = dt

$$y{{e}^{{{\tan }^{-1}}x}}=\frac{1}{2}{{e}^{2{{\tan }^{-1}}x}}+c$$,

Example 2: Solve x (x – 1) dy/dx – y = x³ (x – 1)³

Solution: Given that x (x – 1) dy/dx – y = x³ (x – 1)³

$$\frac{dy}{dx}-\frac{1}{x(x-1)}y={{x}^{2}}{{(x-1)}^{2}}$$,

P = -1/ x (x – 1), Q = x² (x – 1)²

$$\int{\frac{-dx}{x(x-1)}}=\int{\left( \frac{1}{x}-\frac{1}{x-1} \right)dx}$$,

= log x – log (x – 1)

= log x/ (x – 1)

Integration factor (I.F) $$={{e}^{\int{-\frac{dx}{x(x-1)}}}}={{e}^{\log \frac{x}{x-1}}}$$,

= x/ (x – 1)

Solution is y.x/ (x – 1) = ∫x³ (x – 1) dx = ∫ (x⁴ – x³) dx

= x⁵/5 – x⁴/4 + c

The solution is xy/ x – 1 = x⁵/5 – x⁴/4 + c.