Let $y_0$ be the solution for differential equation $y'=ky$, where $y_0(0)=1$. What is the solution $y_1$ of differential equation, when $y_1(a)=b$?
Express function $y_1$ with function $y_0$:
$$y_1(x)= \underline~\underline~\underline~\underline~ \cdot y_0(~\underline~\underline~\underline~\underline~~)$$
I don't know where to start to solve this one, please help?
Note: the empty space "$\underline~\underline~\underline~\underline~$" is for the answer.
The way this problem is written is very confusing.
We have the Separable Equation and can solve
$$\displaystyle \int \dfrac{1}{y}~dy = \int k~dx \implies y(x) = c_1e^{k x}$$
At the first initial condition, it seems like they want to call this solution $y_0(x)$, we have
$$y_0(0) = c_1 e^{k \times 0} = c_1 e^0 = c_1 = 1 \implies y_0(x) = e^{k x}$$
At a second initial condition, it seems like they want to call this solution $y_1(x)$, we have
$$y_1(a) = c_1 e^{k \times a} = b \implies c_1 = be^{-ka} \implies y_1(x) = be^{-ka}e^{k x}$$
Now, we have
$$y_1(x) = be^{-ka}e^{k x} = be^{-ka} y_0(x)$$
Update
It is entirely possible that they wanted this alternate representation
$$y_1(x)=b·y_0(x-a)$$