Question

y′−ay=0,y(x0)=y0
• y=−y0e^−ax0e^−ax
• y=y0 e^ax0 e^−ax
• y=−y0 e^−ax0 e^ax
• y=y0 e^−ax0 e^ax

Answer 1

Example. Find the unique solution of the initial value problem:

$$y' - ay = 0, \quad y(x_0) = y_0.$$

Solution: Rewriting $y' = ay$ we get

$$\frac{y'}{y} = a \Rightarrow \ln(|y|)' = a \Rightarrow \ln(|y|) = ax + c_0.$$

We now compute exponentials on both sides, to get

$$y(x) = \pm e^{ax + c_0} = \pm e^{ax} e^{c_0}, \quad \text{denote } c = \pm e^{c_0}, \quad \text{then } y(x) = c e^{ax},$$

where $c$ is an arbitrary constant. The initial condition determines $c$,

$$y_0 = y(x_0) = c e^{a x_0} \Rightarrow c = y_0 e^{-a x_0}.$$

Then, the unique solution to the initial value problem above is

$$y(x) = y_0 e^{-a x_0} e^{a x} = y_0 e^{a (x - x_0)}.$$