Linear Differential Equation - Practice Questions & MCQ

The linear differential equations are those in which the variable and its derivative occur only in the first degree.

An equation of the form

$
\frac{d y}{d x}+P(x) \cdot y=Q(x)
$

Where P(x) and Q(x) are functions of x only or constant is called a linear equation of the first order.

To solve the differential equation (i)

multiply both sides of Eq (i) by $\int e^{P(x) d x}$, we get
$
\begin{array}{ll} 
& e^{\int P(x) d x}\left(\frac{d y}{d x}+P(x) y\right)=e^{\int P(x) d x} \cdot Q(x) \\
\text { i.e. } & e^{\int P(x) d x} \cdot \frac{d y}{d x}+y \cdot P(x) \frac{d}{d x}\left(e^{\int P(x) d x}\right)=Q e^{\int P(x) d x} \\
\text { or } & \frac{d}{d x}\left(y e^{\int P(x) d x}\right)=e^{\int P(x) d x} \cdot Q(x)
\end{array}
$

Integrating both sides, we get
or $\quad \int \mathrm{d}\left(y \mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}\right)=\int\left(\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}} \cdot \mathrm{Q}(\mathrm{x})\right) \mathrm{dx}$
$
\Rightarrow \quad \mathrm{ye}^{\int P(x) d x}=\int Q(x) e^{\int P(x) d x} d x+C
$

Which is the required solution of the given differential equation.

The term $\mathrm{e}^{\int \mathrm{P}(\mathrm{x}) \mathrm{dx}}$ which converts the left-hand expression of the equation into a perfect differential is called an Integrating factor (IF).

Thus, we remember the solution of the above equation as
$
y(\mathrm{IF})=\int Q(\mathrm{IF}) d x+C
$

NOTE : 

Sometimes a given differential equation becomes linear if we take x as the dependent variable and y as the independent variable.