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I have a rather simple question which I seemingly can't get my head around to solve.

$y_0$ is a solution to differential equation $y'=k*y$, for which $y_0(0)=1$

What is the solution $y_1$, when $y_1(a)=b$?

The answer is supposed to be given with $y_1=?*y_0(?)$

I know there is the so-called order reduction method, but I am not exactly familiar with it yet and assume there's another, simpler way to solve this.

I have just started doing differential equations, so apologies for such a 'simple' question.

And thanks for all help.

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    I am also a beginner but here is my take: it seems like cauchy ode. $$y'=ky \iff\frac{dy}{dx}=ky\iff \frac{dy}{y}=kdx$$ Integrating both side we get: $$ln|y|=xk+c$$ raising to $e$ to get $$y=e^{xk}C$$ Now you are given that y(0)=1 so plug it in to get $$1=e^{0}C$$ So you $C=1$ and in general $$y=e^{xk}$$ – 2017-01-25
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    It seems like the correct answer is $y_1$=$b$+$y_0$(?), but I still need to figure out what value of $y_0$ to use here. This seems like a really odd task. – 2017-01-25

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In this case you could only reduce the order to zero. You can of course try that, it is equivalent in this case to the use of an integrating factor.

Inserting $y=cy_0$ gives $$ y'=cy_0'+c'y_0=ky=cky_0\iff c'y_0=0 $$ That is, you get a solution for any constant $c$.

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    Thanks for the reply, but I am still not sure if I exactly follow here. How is constant $c$ related to the constants $a$ and $b$ given in the task? – 2017-01-25
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    Obviously $cy_0(a)=b$, and given that $y_0(x)=e^{kx}$, $c=e^{-ka}b$. – 2017-01-25