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

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

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

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

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
$y(\mathrm{IF})=\int Q(\mathrm{IF})dx+C$

NOTE : 

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