Substituting $y^2$ in a differential equation

Question

I have the IVP:

$$2yy'+5 = y^2 + 5x; y(0)=6.$$

In order to solve this, I have attempted to make the substitution $u = y^2 \implies \sqrt{u} = y$. However, I am not sure how to solve for $u'$, so I don't know how to proceed in the problem.

I know that $\frac{du}{dx} = \frac{d(y^2)}{dx}$ but I'm not really sure what that means or how to deal with it.

Answer 1

If $u = y^2$, then by the chain rule $$ u' = (y^2)' = 2yy'. $$ From your original equation, you see that $$ 2yy' = y^2+5x-5 = u+5x-5, $$ so that the equation for $u$ becomes $$ u' = u+5x-5. $$ Also take care that the initial condition for $u$ then becomes $$ u(0) = y(0)^2 = 6^2 = 36. $$

Answer 2

Since we have $u'(x) = \frac{d}{dx} y(x)^2 = 2y'(x) y(x)$, we can translate the IVP to $$\tag{1} u'(x) = u(x)+5x-5 \quad \text{with} \quad u(0) = y^2(0) = 36.$$ Note that this approach only works because the intial condition isn't zero. The IVP (1) is an inhomogeneous linear ordinary differential equation of degree one and can be solved as usual.