# Linear Differential Equation

## Linear Differential Equation

A differential equation is said to be linear. If the dependent variable and its differential coefficients occur in the first degree only and are not multiplied together.

Method 1: The most general from of a linear equation of the first order is $$\frac{dy}{dx}+Py=Q$$,

Where, P and Q are any function of x.

To solve such equation, find integrating factor $$IF={{e}^{\int{P.dx}}}$$,

On multiplying the form by $${{e}^{\int{P.dx}}}$$ on both sides, we get

$$\frac{dy}{dx}+Py=Q$$,

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

$$\frac{dy}{dx}.{{e}^{\int{P.dx}}}+P.y.\frac{d}{dx}{{e}^{\int{P.dx}}}=Q.{{e}^{\int{P.dx}}}$$,

$$\frac{d}{dx}\left( y.{{e}^{\int{P.dx}}} \right)=Q.{{e}^{\int{P.dx}}}$$,

Integration applying on both sides

$$\int{\frac{d}{dx}\left( y.{{e}^{\int{P.dx}}} \right)}.dx=\int{Q.{{e}^{\int{P.dx}}}}.dx$$,

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

Which is the required solution of the given differential equation.

Method 2: The most general from of a linear equation of the first order is $$\frac{dx}{dy}+Px=Q$$,

Where, P and Q are any function of y.

To solve such equation, find integrating factor $$IF={{e}^{\int{P.dy}}}$$,

On multiplying the form by  $$IF={{e}^{\int{P.dy}}}$$ on both sides, we get

$$\frac{dx}{dy}+Px=Q$$,

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

$$\frac{dx}{dy}.{{e}^{\int{P.dy}}}+P.x.\frac{d}{dx}{{e}^{\int{P.dy}}}=Q.{{e}^{\int{P.dy}}}$$,

$$\frac{d}{dy}\left( y.{{e}^{\int{P.dy}}} \right)=Q.{{e}^{\int{P.dy}}}$$,

Integration applying on both sides

$$\int{\frac{d}{dy}\left( y.{{e}^{\int{P.dy}}} \right)}.dy=\int{Q.{{e}^{\int{P.dy}}}}.dy$$,

$$x.{{e}^{\int{P.dy}}}=\int{Q.{{e}^{\int{P.dy}}}}+C$$,

Which is the required solution of the given differential equation.