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I have a differential equation:


Here $K>0$ is a fixed constant and $f(x)$ is some (as yet) unknown function of $x$, which is in our hands to chose.

What I want is that I should decide $f(x)$ suitably to find a solution $y(x)$ of the above with the condition $y(0)=1$ where $y(x)$ is a rational function of $x$, i.e. a quotient of two polynomials. The solution should not be free of $K$.

I tried setting $f(x)$ so that the coefficient of $K$ becomes $1$ but the differential equation turned out to be so complicated that I could not solve it.

Can anyone offer any help or suggestions please?

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up vote 1 down vote accepted

Let $R(x)$ be any rational function such that $R(0)=1$ and define $$ f(x)=\Bigl(R''-\frac{3}{2(1+x)(2-x)}R'+\frac{3R}{4(1+x)(2-x)}\Bigr)\frac{2x(1+x)(2-x)}{(1+x)^2R}. $$ The $R$ is a solution of your equation with $K=1$.

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Thanks for your reply. However I am not looking for a solution which is free of $K$, or is true for a fixed value of $K$. – Shahab Oct 11 '12 at 11:34
Then I do not understand your question. – Julián Aguirre Oct 11 '12 at 12:38
Quite simply, I want a solution which is a rational function of $x$ but the term $K$ is present in that solution (eg $(1+Kx)/(1-x)$.) – Shahab Oct 11 '12 at 12:47

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