Ideas for solving this IVP I am curious how to approach solving the initial value problem: $\begin{cases} y'(t) = 5t - 3\sqrt{y} \\ y(0) = 2 \end{cases}$.
The equation isn't separable, and more generally it is not an exact equation.  Nor does it seem to be readily convertible into an exact equation via an integrating factor.  I am interested in obtaining at least an implicit expression for $y$.  Is it possible to use a Laplace transform to solve this nonlinear IVP?  If not, what approach might one take?
How do I solve the ODE by hand, without the help of Maple?
 A: Let $u=\sqrt y$ ,
Then $y=u^2$
$y'=2uu'$
$\therefore2uu'=5t-3u$
$\dfrac{du}{dt}=\dfrac{5t}{2u}-\dfrac{3}{2}$
Luckily this becomes a first-order homogeneous ODE.
Let $v=\dfrac{u}{t}$ ,
Then $u=tv$
$\dfrac{du}{dt}=t\dfrac{dv}{dt}+v$
$\therefore t\dfrac{dv}{dt}+v=\dfrac{5}{2v}-\dfrac{3}{2}$
$t\dfrac{dv}{dt}=-\dfrac{2v^2+3v-5}{2v}$
$\dfrac{2v}{2v^2+3v-5}~dv=-\dfrac{dt}{t}$
$\int\dfrac{2v}{(2v+5)(v-1)}~dv=-\int\dfrac{dt}{t}$
$\int\left(\dfrac{10}{7(2v+5)}+\dfrac{2}{7(v-1)}\right)~dv=-\int\dfrac{dt}{t}$
$\dfrac{5\ln(2v+5)}{7}+\dfrac{2\ln(v-1)}{7}=-\ln t+c_1$
$5\ln(2v+5)+2\ln(v-1)=c_2-7\ln t$
$\ln((2v+5)^5(v-1)^2)=c_2-\ln t^7$
$(2v+5)^5(v-1)^2=\dfrac{C}{t^7}$
$\left(\dfrac{2u}{t}+5\right)^5\left(\dfrac{u}{t}-1\right)^2=\dfrac{C}{t^7}$
$(2u+5t)^5(u-t)^2=C$
$(2\sqrt y+5t)^5(\sqrt y-t)^2=C$
$y(0)=2$ :
$C=256\sqrt2$
$\therefore(2\sqrt y+5t)^5(\sqrt y-t)^2=256\sqrt2$
A: Wolfram Alpha identifies it as an instance of Chini's equation, see here.
A: Let us try the substitution
$$y=z^\alpha t^\beta.$$
$$\alpha z^{\alpha-1}z't^{\beta}+\beta z^\alpha t^{\beta-1}=5t-3z^{\alpha/2}t^{\beta/2}.$$
With $\beta=2$, three terms can be grouped,
$$\alpha z^{\alpha-1}z't^2+2z^\alpha t=5t-3z^{\alpha/2}t,$$
or
$$\alpha z^{\alpha-1}z'=\frac{5-3z^{\alpha/2}-2z^\alpha}t,$$ which is separable.
Setting $\alpha=2$ to avoid fractional exponents,
$$\frac{2zz'}{5-3z-2z^2}=\frac1t.$$
After integration of the rational fractions,
$$\frac17(-2\log(3-3z)-5\log(6z+15))=\log(Ct),$$
$$(3t-3\sqrt y)^2(6\sqrt y+15t)^5=C.$$
I am afraid there is nothing more that you can do to make $y$ or $t$ an explicit function of the other variable. Anyway, you can build a parametric form, starting from
$$u^2v^5=C,$$giving the linear system
$$3t-3\sqrt y=u,\\
6\sqrt y+15t=v=\sqrt{\frac C{u^5}}.$$
You solve for $y$ and $t$ as a function of $u$.
$$t=\frac1{21}\left(\sqrt{\frac C{u^5}}+2u\right),\\
y=\frac1{21^2}\left(\sqrt{\frac C{u^5}}-5u\right)^2,\\
C=3^72^8\sqrt2.$$
