Question

𝑦^1/2 𝑑𝑦/𝑑𝑥+𝑦3/2=1; 𝑦(0)=4

Answer 1

Given: $y^{\frac{1}{2}}\frac{dy}{dx}+y^{\frac{3}{2}}=1$

Require to solve the given differential equation.

Now $y^{\frac{1}{2}}\frac{dy}{dx}+y^{\frac{3}{2}}=1\Rightarrow y^{\frac{1}{2}}\frac{dy}{dx}=1-y^{\frac{3}{2}}$

$\Rightarrow \frac{dy}{dx}=\frac{1-y^{\frac{3}{2}}}{y^{\frac{1}{2}}}$

By using separation of variables, we get

$\frac{y^{1/2}}{1-y^{3/2}}dy=dx$

Intetrating on both sides, we get

$\int \frac{y^{1/2}}{1-y^{3/2}}dy=\int dx$

To evaluate the above integral, let us take the substitution

$u=1-y^{3/2}\Rightarrow du=[0-\frac{3}{2}y^{1/2}]dy\Rightarrow du=-\frac{3}{2}y^{1/2}dy$

$\Rightarrow y^{1/2}dy=-\frac{2}{3}du$

Then the integral becomes

$\int \frac{-\frac{2}{3}du}{u}=\int dx$

$-\frac{2}{3}ln\left | u \right |=x+c$

Substituting $u=1-y^{3/2}$ , we get

$-\frac{2}{3}ln\left | 1-y^{3/2} \right |=x+c$

Use the initial condition $y(0)=4$ to find the value of c.

$y(0)=4\Rightarrow -\frac{2}{3}ln\left | 1-4^{3/2} \right |=0+c\Rightarrow c=-\frac{2}{3}ln(7)$

Then, solution to the given differential equation is

$-\frac{2}{3}ln\left | 1-y^{3/2} \right |=x-\frac{2}{3}ln(7)$

Therefore general solution is

$\frac{2}{3}ln\left | \frac{1-y^{3/2}}{7} \right |=-x$