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$$\begin{array}{rl}\frac{dy}{dt}+y^{\frac{3}{2}}\left(t\right)& =a^{\frac{3}{2}}\\ y\left(0\right)& =0\end{array}$$

Write as

$$\begin{array}{rl}(y^{\frac{3}{2}}-a^{\frac{3}{2}})dt+dy& =0\\ \text{(1)}& M(t,y)dt+N(t,y)dy& =0\end{array}$$

Where

$$\begin{array}{rl}M& =y^{\frac{3}{2}}-a^{\frac{3}{2}}\\ N& =1\end{array}$$

Check if exact

$$\begin{array}{rl}\frac{\partial M(t,y)}{\partial y}& =\frac{3}{2}{y}^{\frac{1}{2}}\\ \frac{\partial N(t,y)}{\partial t}& =0\end{array}$$

Since $\frac{\partial M(t,y)}{\partial y}\ne \frac{\partial N(t,y)}{\partial t}$ then Not exact. Trying integrating factor $A=\frac{\frac{\partial N}{\partial t}-\frac{\partial M}{\partial y}}{M}=\frac{-\frac{3}{2}{y}^{\frac{1}{2}}}{y^{\frac{3}{2}}-a^{\frac{3}{2}}}$, Since it is a function of $y$ alone, then it (1) can be made exact. The integrating factor is

$$\begin{array}{rl}\mu & =e^{\int Ady}\\ & =e^{\int \frac{-\frac{3}{2}{y}^{\frac{1}{2}}}{y^{\frac{3}{2}}-a^{\frac{3}{2}}}dy}\\ & =e^{-\ln (a^{\frac{3}{2}}-y^{\frac{3}{2}})}\\ & =\frac{1}{a^{\frac{3}{2}}-y^{\frac{3}{2}}}\end{array}$$

Multiplying (1) by this integrating factor, now it becomes exact

$$\mu M(t,y)dt+\mu N(t,y)dy=0$$

Now we follow standard method for solving exact ODE. Let

$$\begin{array}{}\text{(2)}& \frac{dU}{dt}& =\mu M=\frac{y^{\frac{3}{2}}-a^{\frac{3}{2}}}{a^{\frac{3}{2}}-y^{\frac{3}{2}}}=-1\text{(3)}& \frac{dU}{dy}& =\mu N=\frac{1}{a^{\frac{3}{2}}-y^{\frac{3}{2}}}\end{array}$$

From (2)

$$\begin{array}{rl}U& =-\int dt\\ \text{(4)}& & =-t+f\left(y\right)\end{array}$$

Substituting this into (3) to solve for $f\left(y\right)$

$$\begin{array}{rl}{f}^{\prime }\left(y\right)& =\frac{1}{a^{\frac{3}{2}}-y^{\frac{3}{2}}}\\ f\left(y\right)& =\frac{-2\sqrt{3}}{3\sqrt{a}}\arctan \left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)-\frac{2}{3\sqrt{a}}\ln (\sqrt{a}-\sqrt{y})+\frac{1}{3\sqrt{a}}\ln (a+\sqrt{ay}+y)+C\end{array}$$

Hence the solution from (4) is

$$U=-t+\frac{-2\sqrt{3}}{3\sqrt{a}}\arctan \left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)-\frac{2}{3\sqrt{a}}\ln (\sqrt{a}-\sqrt{y})+\frac{1}{3\sqrt{a}}\ln (a+\sqrt{ay}+y)+C$$

But $\frac{dU}{dt}=0$, hence $U=C_1$. Therefore, collecting constants into one, the solution is (implicit form)

$$t+\frac{2\sqrt{3}}{3\sqrt{a}}\arctan \left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}\ln (\sqrt{a}-\sqrt{y})-\frac{1}{3\sqrt{a}}\ln (a+\sqrt{ay}+y)=C$$

From initial conditions

$$\begin{array}{rl}\frac{2\sqrt{3}}{3\sqrt{a}}\arctan \left(\frac{1}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}\ln \left(\sqrt{a}\right)-\frac{1}{3\sqrt{a}}\ln \left(a\right)& =C\\ C& =\frac{2\sqrt{3}}{3\sqrt{a}}\frac{\pi }{6}+\frac{2}{3\sqrt{a}}\ln \left(\sqrt{a}\right)-\frac{1}{3\sqrt{a}}\ln \left(a\right)\\ C& =\frac{2\sqrt{3}}{3\sqrt{a}}\frac{\pi }{6}\\ C& =\frac{\pi \sqrt{3}}{9\sqrt{a}}\end{array}$$

Hence final solution for $y\left(t\right)$ in implicit form is

$$\begin{array}{rl}t+\frac{2\sqrt{3}}{3\sqrt{a}}\arctan \left(\frac{1+2\sqrt{\frac{y}{a}}}{\sqrt{3}}\right)+\frac{2}{3\sqrt{a}}\ln (\sqrt{a}-\sqrt{y})-\frac{1}{3\sqrt{a}}\ln (a+\sqrt{ay}+y)& =\frac{\pi \sqrt{3}}{9\sqrt{a}}\\ 3t\sqrt{a}+2\sqrt{3}\arctan \left(\frac{\sqrt{a}+2\sqrt{y}}{\sqrt{3}\sqrt{a}}\right)+6\ln (\sqrt{a}-\sqrt{y})-\ln (a+\sqrt{ay}+y)& =\frac{\pi \sqrt{3}}{3}\end{array}$$