Question

Write a differential equation in wich \( y^2=4(t+1) \) is a solutiion.

Answer 1

Solution

we have \( y^2=4(t+1)\quad (1) \)

\( y^2=4t+4\implies 2y\frac{dy}{dt}=4\implies y\frac{dy}{dt}=2 \)

\( \implies y'=\frac{2}{y}\iff \frac{dy}{dt}=\frac{2}{y}\iff y=dy=2dt \)

\( \iff \int ydy=2\int dt\implies \frac{1}{2}y^2+c_1=2t+c_2 \)

\( \iff y^2=4t+2(c_2-c_1)\implies y^2=4t+c\quad (2) \)

\( \implies\begin{cases} y^2(0)=c\quad (2) & \quad \\ y^2(0)=4\quad (1) & \quad \end{cases}\implies c=4 \)

Then \( y'=4t+4 \) is a soltion of  \( y^2-\frac{2}{y}=0 \)

Therefore, the differential equation is \( y'-\frac{2}{y}=0 \)