Chapter 2, Problem 1MP

To determine

The solution of the differential equation $\frac{dy}{dx}=\frac{x^3-2y}{x}$.

Answer to Problem 1MP

The solution of the given differential equation is $\underset{\_}{y=\frac{x^3}{5}+\frac{c}{x^2}}$.

Explanation of Solution

Definition used:

The first order linear equation is of the form $\frac{dy}{dt}+p\left(t\right)y=q\left(t\right)$ where p and q are function of the independent variable t.

Formula used:

The integrating factor $\mu \left(t\right)$ of the first order differential equation $\frac{dy}{dt}+p\left(t\right)y=q\left(t\right)$ is given as $\mu \left(t\right)=e^{{\displaystyle \int p\left(t\right)}dt}$ and the solution of the differential equation is given by $y=\frac{1}{\mu \left(t\right)}\left[{\displaystyle \int \mu \left(s\right)q\left(s\right)ds}+c\right]$.

Calculation:

The given differential equation is $\frac{dy}{dx}=\frac{x^3-2y}{x}$.

Rewrite the given differential equation of the form $\frac{dy}{dt}+p\left(t\right)y=q\left(t\right)$ as,

$\begin{array}{l}\frac{dy}{dx}=\frac{x^3-2y}{x}\\ \frac{dy}{dx}=\frac{x^3}{x}-\frac{2y}{x}\\ \frac{dy}{dx}+\frac{2y}{x}=x^2\\ \frac{dy}{dx}+\left(\frac{2}{x}\right)y=x^2\end{array}$

Note that, $p\left(t\right)=\frac{2}{x}\text{ and }q\left(t\right)=x^2$.

Compare with the formula and obtain the integrating factor of the differential equation.

$\begin{array}{c}\mu \left(x\right)=e^{{\displaystyle \int \frac{2}{x}}dx}\\ =e^{2{\displaystyle \int \frac{1}{x}}dx}\\ =e^{2\left(\log x\right)}\left(\because {\displaystyle \int \frac{1}{x}}dx=\log x\right)\\ =e^{\left(\log x\right)}{}^{{}^2}\\ =x^2\end{array}$

Now, obtain the solution as follows.

$\begin{array}{c}y=\frac{1}{\mu \left(x\right)}\left[x+c\right]\\ =\frac{1}{x^2}\left[{\displaystyle \int x^2\left(x^2\right)dx}+c\right]\\ =\frac{1}{x^2}\left[{\displaystyle \int x^{4}dx}+c\right]\\ =\frac{1}{x^2}\left[\frac{x^5}{5}+c\right]\\ =\frac{x^3}{5}+\frac{c}{x^2}\end{array}$

Therefore, the solution of the given differential equation is $\underset{\_}{y=\frac{x^3}{5}+\frac{c}{x^2}}$.