# How to obtain coefficients after solving differential equation of order 2 with initial condition?

I have the following answer of a differential equation of order two

y[x]=C[1] (2/5)^(2/5) Sqrt[1 + x] \[Lambda]^(1/5)
BesselJ[-(2/5), 4/5 (1 + x)^(5/4) Sqrt[\[Lambda]]]  Gamma[3/
5] + C[2] (2/5)^(2/5) Sqrt[1 + x] \[Lambda]^(1/5)
BesselJ[2/5, 4/5 (1 + x)^(5/4) Sqrt[\[Lambda]]]  Gamma[7/5]


I want to obtain C[1] and C[2] with the following initial conditions:

y[-1] = 0 and D[y[-1]] = 1.

Can anyone help me?

• When solving differential equations, you should be able to include your initial conditions (i.e. DSolve[{y'[x] + y[x] == a Sin[x], y[0] == 0}, y[x], x]). Are you required to solve it by first getting the general solution?
– Asas
Apr 22, 2015 at 6:30
• In the first I have integro differential equation. Apr 22, 2015 at 6:47

If you redefine your function as y[x_] = ... rather than y[x] = ..., then you can do the following:
Solve[{Limit[y[x], x -> -1] == 0, Limit[D[y[x], x], x -> -1] == 1}, {C[1], C[2]}]

$$\left\{\left\{c_1\to 0,c_2\to \frac{84 \sqrt[5]{\frac{2}{5}} \Gamma \left(\frac{2}{5}\right)}{25 \lambda ^{2/5} \Gamma \left(\frac{17}{5}\right)}\right\}\right\}$$ $$\frac{84 \left(\frac{2}{5}\right)^{3/5} \sqrt{x+1} \Gamma \left(\frac{2}{5}\right) \Gamma \left(\frac{7}{5}\right) J_{\frac{2}{5}}\left(\frac{4}{5} (x+1)^{5/4} \sqrt{\lambda }\right)}{25 \sqrt[5]{\lambda } \Gamma \left(\frac{17}{5}\right)}$$
Note that you have to use the Limit commands in the Solve, since the Bessel function of negative order diverge as their arguments go to zero; since the first term of your general solution contains $\sqrt{1+x} J_{-2/5}((1+x)^{5/4})$, it's indeterminate at $x = -1$.