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I need help with this exercise.

Solve the differential equation $$y(2t^{2}\sqrt{y}+2)\,\mathrm dt + t(t^{2}\sqrt y +2) \,\mathrm dy=0$$ using $u = t^{2}\sqrt y$.

My steps: $$\frac{\mathrm du}{\mathrm dt} = 2t \sqrt y + t^{2}\frac{1}{2 \sqrt y} y'$$ And $y= \frac{u}{t^2}$ gives $$y'= \frac{1}{t^2}\times (- \sqrt u + \frac{1}{2 \sqrt u})$$ When I replacement in the equation, the solution is not real number, I tried this exercise by different forms.

Thanks if you give me the solution. Also, please I need some recommendations about find interesting books or material with differential equation by change of variables.

ultralegend5385
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Estre
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1 Answers1

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Your first step is correct: $$ \frac{du}{dt}=2t\sqrt{y}+\frac{t^2y'}{2\sqrt{y}}. \tag{1} $$ Now, let's substitute $$ y'=-\frac{y(2t^2\sqrt{y}+2)}{t(t^2\sqrt{y}+2)} \tag{2} $$ in $(1)$: $$ \frac{du}{dt}=2t\sqrt{y}-\frac{t\sqrt{y}(2t^2\sqrt{y}+2)}{2(t^2\sqrt{y}+2)}. \tag{3} $$ Substituting $\sqrt{y}=\frac{u}{t^2}$ in $(3)$, we obtain $$ \frac{du}{dt}=\frac{2u}{t}-\frac{u(2u+2)}{2t(u+2)}=\frac{u(u+3)}{t(u+2)}. \tag{4} $$ Equation $(4)$ is a separable ODE. You should have no difficulty in solving it.

Remark. Using the fact that the original differential equation is exact, one can find its solution in a less convoluted way. The result is $$ \frac{2}{3}t^3y^{3/2}+2ty=C. \tag{5} $$

Gonçalo
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