Originally Posted by

**Chris L T521** It all is correct so far. Just keep in mind that y is the subject of the equation. So we can rewrite what you have as $\displaystyle y^2+3y=t+C$. Note that if you complete the square, you end up with $\displaystyle \left(y+\tfrac{3}{2}\right)^2=t+C-\tfrac{3}{2}\implies \left(y+\tfrac{3}{2}\right)^2=t+K$, where $\displaystyle K$ is another constant.

Now, applying the initial condition $\displaystyle y(0)=1$, we have $\displaystyle \tfrac{25}{4}=K$

Therefore, $\displaystyle \left(y+\tfrac{3}{2}\right)^2=t+\tfrac{25}{4}$

Does this make sense?