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I bit further away along the pages we will solve the equation

You can have a look at it here. The problem we quickly encountered was that the solution is a non-elementary function. Let us tackle this in another way. Previously, in the section on power series, we were able to solve the equation *y*´=*y* by assuming the solution could be expressed as a power series. Let us attempt something similar here.

Say we want a solution that goes through the point (1, 2). So we start by assuming we have a solution that can be expressed with a Taylor series, i.e. that

In this case, we have that

and that

so

Differentiating our differential equation once gives us

and thus

Differentiate again

so

Differentiate again

So, an approximate solution should be

In the graph below we can see the slope field, the particular solution in green (found on the above-mentioned page) and the approximate solution in violet.

If we try that for *x*=4 we get *y*=12.5, which is rather far from the about 13.96 we should get according to the exact solution. If we add some more terms we get

From this point on we have that

This gives us the sequence 2, 3, 7, 15, 36, 96, 267, 852 and so on. We can use these values to get a better approximation. Using our exact method we get 13.96225, using 15 terms in our series expansion we get 13.69221, so quite close.

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