**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.