Question

Find the general solution for the following differential equation, $x\dfrac{{dy}}{{dx}} - y = 2{x^3}$?

Answer 1

Here we need to find the general solution. If we observe the given differential equation and on further simplification it is of the form $\dfrac{{dy}}{{dx}} + Py = Q$. The general solution is given by $y.\left( {I.F} \right) = \int {Q.(I.F)dx + c}$, where I.F is the integrating factor.

Complete step-by-step solution:
Given,
$x\dfrac{{dy}}{{dx}} - y = 2{x^3}$
Divide the whole D.E by ‘x’ we have,
$\dfrac{{dy}}{{dx}} - \dfrac{y}{x} = \dfrac{{2{x^3}}}{x}$
$\dfrac{{dy}}{{dx}} - \dfrac{y}{x} = 2{x^2}$
On comparing with $\dfrac{{dy}}{{dx}} + Py = Q$, we have,
$P = - \dfrac{1}{x}$ and $Q = 2{x^2}$.
Now we need to find the integrating factor,
$I.F = {e^{\int {Pdx} }}$
On substituting we have,
$I.F = {e^{\int { - \dfrac{1}{x}dx} }}$
On integrating we have,
$I.F = {e^{ - \log x}}$
$I.F = {e^{\log {x^{ - 1}}}}$
On applying logarithm power rule,
$I.F = {e^{\log \dfrac{1}{x}}}$
Thus we have,
$\Rightarrow I.F = \dfrac{1}{x}$
We know the general solution is
$y.\left( {I.F} \right) = \int {Q.(I.F)dx + c}$
Then
$y.\left( {\dfrac{1}{x}} \right) = \int {2{x^2}.\left( {\dfrac{1}{x}} \right)dx + c}$
$y.\left( {\dfrac{1}{x}} \right) = \int {2x.dx + c}$
$\dfrac{y}{x} = 2\dfrac{{{x^2}}}{2} + c$
multiply the whole equation by x we have,
$\Rightarrow y = {x^3} + cx$. This is the general solution and ‘c’ is the integration constant.

Thus the required answer is $y = {x^3} + cx$.

Note: We know that the integration of ${x^n}$ with respect to ‘x’ is $\int {{x^n}.dx} = \dfrac{{{x^{n + 1}}}}{{n + 1}} + c$, where ‘c’ is the integration constant. In case of indefinite integral we will have integration constant and in definite integral we will have upper limit and lower limit hence we will not have integration content in case of definite integral.