# solving differential equations with function coefficients using Laplace Transform

Does there exits a method to solve an $n$-th order liner differential equation with "function coefficients" using Laplace transform. It is well known that the identity $$L\left\{ {{t^n}f\left( t \right)} \right\} = {\left( { - 1} \right)^n}\frac{{{d^n}}}{{d{s^n}}}F\left( s \right),$$ where $L\left\{ {{t^n}f\left( t \right)} \right\}$ is the Laplace transform of $t^n f\left( t \right)$, can be used to solve such problems. However, it is not easy and indeed we didn't make any transform (we just transform D.E. with function coefficients to one another). So my Question: Does there a direct method (Like D.Es with constant coefficients) to solve this type of D.E.?

• there is a method when the coefficients are linear functions, see math.stackexchange.com/questions/816129/… – user8268 Jun 5 '14 at 11:37
• Using the substitution $x = e^t$ the Cauchy-Euler equations of the form $a x^2 y'' + b x y' + cy = h(x)$ where $a,b,c$ constants can be transformed into a 2nd order DE with constant coefficient $a y'' + (b-a) y' + cy = h(e^t)$. My question how about other types?! – mwomath Jun 9 '14 at 9:42
• e.g., how about $g(x) y'' + 4y = 0$ We may think about the analyticity of the leading function in the DE (say about $x =0$) by setting $g(x)=\sum\limits_{k = 0}^\infty {{a_k}{x^k}}$. – mwomath Jun 9 '14 at 9:55

Let $y=\int_Ce^{xs}K(s)~ds$ , then you can obtain the first-order ODE for $K(s)$ . Once you find $K(s)$ , you can substitute back to $\int_Ce^{xs}K(s)~ds$ . Since the procedure in fact suitable for any complex number $s$ , you can rewrite the integral form as $\int_{a_n}^{b_n}e^{x(p_n+q_ni)t}K((p_n+q_ni)t)~d((p_n+q_ni)t)$ . By choosing suitable $x$-independent real numbers groups of $a_n$ , $b_n$ , $p_n$ and $q_n$ , you can find the linear independent solutions in integral forms.
Sometimes you need to find the linear independent solutions with the combinations of different types of integral forms , for example in Particular solution to a Riccati equation $y' = 1 + 2y + xy^2$.