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?
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]}]
yICs[x_] = (y[x] /. %)
$$ \left\{\left\{c_1\to 0,c_2\to \frac{84 \sqrt[5]{\frac{2}{5}} \Gamma \left(\frac{2}{5}\right)}{25 \lambda ^{2/5} \Gamma \left(\frac{17}{5}\right)}\right\}\right\} $$ $$ \frac{84 \left(\frac{2}{5}\right)^{3/5} \sqrt{x+1} \Gamma \left(\frac{2}{5}\right) \Gamma \left(\frac{7}{5}\right) J_{\frac{2}{5}}\left(\frac{4}{5} (x+1)^{5/4} \sqrt{\lambda }\right)}{25 \sqrt[5]{\lambda } \Gamma \left(\frac{17}{5}\right)} $$
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 $x = -1$.
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? $\endgroup$