How to solve $y'=ky$ by $y_1(x)= \underline~\underline~\underline~\underline~ \cdot y_0(~\underline~\underline~\underline~\underline~~)$ 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

$$y_1(x)=b·y_0(x-a)$$