I am trying to solve equation symoblicaly with Mupad, where on the left side I have differential of $p_3$, and on the right side I also have normal value of $p_3$ (not differential).
o := ode(p3'(z) = (-diff(p1,z)*p2-diff(p2,z)*p1-8*diff(p2,z)-8*diff(p1,z)/p0+8*p1*diff(p0,z)/p0^2-diff(p0,z)*p3(z))/p0,p3(z))
On the other side variables $p_0$,$p_1$,$p_2$ are:
p0:=sqrt(1+64*(1-z))
p1:=8*(1/sqrt(1+64*(1-z))-1)
p2:=(-ln(sqrt(1+64*(1-z)))+4*(1-1/(1+64*(1-z))))/sqrt((1+64*(1-z)))
When I gave command
solve(o)
to Mupad, I have got some integral which is not solved, it is not final result. How can I do this with Mupad, and is it possible so solve it symbolicaly on this way? Or is it Wolfram Mathematica better and would it solve it, or there is some other problem?
Equations: $$p_0=(1+64\cdot(1-z))^{0.5}$$ $$p_1=8 \cdot(\dfrac{1}{p_0}-1)=8 \cdot(\dfrac{1}{(1+64\cdot(1-z))^{0.5}}-1)$$ $$p_2=\dfrac{8}{p_0}(-ln(p_0)+4\cdot(1-\dfrac{1}{p_0^2}))$$ $$p_2=\dfrac{8}{(1+64\cdot(1-z))^{0.5}}(-ln((1+64\cdot(1-z))^{0.5})+4\cdot(1-\dfrac{1}{(1+64\cdot(1-z))}))$$
Differential equation which I am solving here: $$p_3'=\dfrac{-p_1'p_2-p_1p_2'-8p_2'-8\dfrac{p_1'}{p_0}+8\dfrac{p_1p_0'}{p_0^2}-p_0'p_3}{p_0}$$
or in other form:
$$\dfrac{dp_3}{dz}=\dfrac{-\dfrac{dp_1}{dz}p_2-p_1\dfrac{dp_2}{dz}-8\dfrac{dp_2}{dz}-8\dfrac{\dfrac{dp_1}{dz}}{p_0}+8\dfrac{p_1\dfrac{dp_0}{dz}}{p_0^2}-\dfrac{dp_0}{dz}p_3}{p_0}$$
Assuming I converted the equations to Mathematica correctly,
$ \text{p3}(z)=\frac{c_1 (65-64 z)^2+64 (65-64 z)^{3/2}+96 (64 z-65)+4 (65-64 z) \log (65-64 z)+32}{(65-64 z)^{5/2}} $