Question

Solve the given differential equation by using appropriate Substitution;

${\mathit{y}}^{\frac{\mathbf{1}}{\mathbf{2}}}\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{+}{\mathit{y}}^{\frac{\mathbf{3}}{\mathbf{2}}}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{4}$

Short answer

We will solve the given differential equation by substituting.

Step-by-step solution

definition of solving differential equation by substitution

Often the first step in solving the differential equation consists of transforming it into another differential equation by means of a substantial.

For example: Suppose we wish to transform the first order differential equation dy/dx=f(x,y) by the substitution y=g(x,u), where u is regarded as a function of the variable x.

solving the following by substitution;

Substituting by u:

$\begin{array}{rcl}u& =& y^{\frac{2}{3}}\\ \frac{du}{dy}& =& \frac{3}{2}{y}^{\frac{1}{2}}\\ \frac{dy}{du}& =& \frac{2}{3y^{{\displaystyle \frac{1}{2}}}}\\ \frac{dy}{dx}& =& \frac{dy}{du}\frac{du}{dx}\\ \frac{dy}{dx}& =& \frac{2}{3y^{{\displaystyle \frac{1}{2}}}}\frac{du}{dx}\\ y^{\frac{1}{2}}\left[\frac{2}{3y^{{\displaystyle \frac{1}{2}}}}\frac{du}{dx}\right]+u& =& 1\\ \frac{2}{3}\frac{du}{dx}+u& =& 1\\ \frac{du}{dx}+\frac{3}{2}u& =& \frac{3}{2}\end{array}$

finding integral factor

$\begin{array}{rcl}{e}^{\int \frac{3}{2}dx}& =& e^{\frac{3}{2}x}\\ e^{\frac{3}{2}x}\left[\frac{du}{dx}+\frac{3}{2}u\right]& =& e^{\frac{3}{2}x}\left[\frac{3}{2}\right]\\ \frac{d}{du}\left[e^{\frac{3}{2}x}u\right]& =& \frac{3}{2}{e}^{\frac{3}{2}x}\\ \int d\left[e^{\frac{3}{2}x}u\right]& =& \int \left[\frac{3}{2}{e}^{\frac{3}{2}x}dx\right]\\ e^{\frac{3}{2}x}u& =& e^{\frac{3}{2}x}+C\\ u& =& 1+C{e}^{-\frac{3}{2}x}\\ y^{\frac{3}{2}}& =& 1+C{e}^{-\frac{3}{2}x}\\ 8& =& 1+C\\ C& =& 7\\ y^{\frac{3}{2}}& =& 1+7e^{-\frac{3}{2}x}\end{array}$

Hence the final answer is$y^{\frac{3}{2}}=1+7e^{-\frac{3}{2}x}$