How to find the root of a differential equation when it is exponential?

Question

I am trying to find the solution to the differential equation $y''+3e^{3x}=0$ given boundary conditions $y(0)=0$ and $y(1)=-1$.

I am supposed to use the bisection method to find the root and the shooting method to solve. However... Since it is an exponential function, it doesn't have a root.

Is there something I'm misunderstanding here, mathematically?

Answer 1

From

$y''(x)=-3 \exp(3x)$ we get $y'(x)=- \exp(3x)+c_1$, hence

$$y(x)= - \frac{1}{3} \exp(3x)+c_1x+c_2.$$

Now use $y(0)=0$ and $y(1)=-1$ to determine $c_1$ and $c_2$.

Answer 2

If you consider what Fred answered $$y(x)=-\frac 13 e^{3x}+c_1x+c_2$$ $$y'(x)=-e^{3x}+c_1$$ The first derivative cancels at $x_*=\frac{\log (c_1)}{3}$. If $y(x_*)>0$, then you have a root somewhere between $x_*$ and $1$.