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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.

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    Is that exactly as it is written? – 2017-01-29
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    No, the original one is in finnish, i translated it. – 2017-01-29
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    Something is not lining up - please reread the question and see if the translation fits. – 2017-01-29
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    Is it better now? – 2017-01-29
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    Since $y_0(0) = 1$, does that mean that $y_1(a) = b$ means $y_1(1) = b$, that is $a = 1$? – 2017-01-29
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    Good point, but I don't know. The question is so confusing for me. – 2017-01-29
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    Duplicate of https://math.stackexchange.com/questions/2113573/given-one-solution-to-a-differential-equation-find-another-solution – 2017-01-29

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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

$$y_1(x)=b·y_0(x-a)$$

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    Thank you so much! :) Now I understand it better. – 2017-01-29
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    What they probably wanted to see inserted was $y_1(x)=b·y_0(x-a)$. – 2017-01-29
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    @LutzL: That could be right too - the problem was translated from Finnish - so it was not at all clear, but I will add that. Thanks! – 2017-01-29
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    Both Moo and LutzL answers are correct, but of course the stack quiz prefers simpler answer...Thanks to both of you, for providing help :) And sorry for my poor english. – 2017-01-29
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    You do not even have to solve the equation. It is linear homogeneous, thus $c·y_0(x)$ is a solution. It is autonomous, thus also $c·y_0(x+d)$ is also a solution. Now adapt parameters,... – 2017-01-29