Specific differential equation with initial conditions

Question

I have diff. equation : $ y'' +2y' = (y')^2 e ^x , y(0)=3, y'(0)=1 $, and i have problem with solving that. I used substitution $ u(x)=y' $ and i got bernoulli's diff. equation. I solved that and got $ y'=[(1/2)*e^x +D*e^{2y'}]^{-1} $ and that is basically confounded differential equation. Any help ?

Answer 1

If $u=y'$, the problem becames $$\begin{cases} u'+2u=u^2 \operatorname {e}^x \\ u(0)=1 \end{cases} $$ which, as you said, is indeed a Bernoulli equation. If we divided by $u^2$ we get $$ \frac{u'}{u^2} + \frac{2}{u} = \operatorname e^{x} $$ and now, after the substitution $v=\frac{1}{u}$, we obtain $$ v'-2v=-\mathrm e^{x} $$

That has solution $$ v(x) = c\operatorname e^{2x}+\operatorname e^{x}=\operatorname e^{x} \left(c\cdot \operatorname e^x +1\right) $$

Therefore $$ u(x)= \frac{1}{\operatorname e^{x} \left(c\cdot \operatorname e^x +1\right)} = \frac{\operatorname e^{-x}}{c\cdot \operatorname e^x +1} $$

Now, if we impose $u(0)=1$, we found

$$ u(0) = \frac{1}{c+1}=1\implies c = 0 $$

Hence the solution is $u(x)=\operatorname{e}^{-x}$.

Finally

$$ y(x)=-\operatorname{e}^{-x}+c $$

And, imposing $y(0)=3$, we have

$$ y(x)=4-\operatorname{e}^{-x} $$