Ideas for solving this IVP

Question

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?

Answer 1

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 $u=tv$

$\dfrac{du}{dt}=t\dfrac{dv}{dt}+v$

$\therefore t\dfrac{dv}{dt}+v=\dfrac{5}{2v}-\dfrac{3}{2}$

$t\dfrac{dv}{dt}=-\dfrac{2v^2+3v-5}{2v}$

$\dfrac{2v}{2v^2+3v-5}~dv=-\dfrac{dt}{t}$

$\int\dfrac{2v}{(2v+5)(v-1)}~dv=-\int\dfrac{dt}{t}$

$\int\left(\dfrac{10}{7(2v+5)}+\dfrac{2}{7(v-1)}\right)~dv=-\int\dfrac{dt}{t}$

$\dfrac{5\ln(2v+5)}{7}+\dfrac{2\ln(v-1)}{7}=-\ln t+c_1$

$5\ln(2v+5)+2\ln(v-1)=c_2-7\ln t$

$\ln((2v+5)^5(v-1)^2)=c_2-\ln t^7$

$(2v+5)^5(v-1)^2=\dfrac{C}{t^7}$

$\left(\dfrac{2u}{t}+5\right)^5\left(\dfrac{u}{t}-1\right)^2=\dfrac{C}{t^7}$

$(2u+5t)^5(u-t)^2=C$

$(2\sqrt y+5t)^5(\sqrt y-t)^2=C$

$y(0)=2$ :

$C=256\sqrt2$

$\therefore(2\sqrt y+5t)^5(\sqrt y-t)^2=256\sqrt2$

Answer 2

Wolfram Alpha identifies it as an instance of Chini's equation, see here.

Answer 3

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 $u$.

$$t=\frac1{21}\left(\sqrt{\frac C{u^5}}+2u\right),\\ y=\frac1{21^2}\left(\sqrt{\frac C{u^5}}-5u\right)^2,\\ C=3^72^8\sqrt2.$$