Question

Verify that the given function is a solution to the differential equation:

y''(x)= 5/y^3 , y = sqrt(x^2 + 5)

Answer 1

I've differentiated twice and so far I've ended up with;

y'' = -x^2(x^2 + 5)^(-3/2) + (x^2 + 5 )^(-1/2)

Answer 2

It worked for me. I will continue from where you stopped:
\[{-x^2 \over (x^2+5)^{3/2}}+{1 \over (x^2+5)^{1/2}}={-x^2+x^2+5 \over (x^2+5)^{3/2}}={5 \over ((x^2+5)^{1/2})^3}\]
But \((x^2+5)^{1/2}=y\) So, \(y''=\frac{5}{y^3}\).

Answer 3

\[y=\sqrt{x^2+5}=(x^2+5)^{1/2}\]
\[y' = (1/2)(2x)(x^2+5)^{-1/2}=x(x^2+5)^{-1/2}\]
\[y'' = (1)(x^2+5)^{-1/2} + (x)(-1/2)(2x)(x^2+5)^{-3/2}\]
\[y'' = (x^2+5)^{-1/2} - x^2(x^2+5)^{-3/2}\]
\[y''=(x^2+5)^1(x^2+5)^{-3/2}-x^2(x^2+5)^{-3/2}\]
\[y''=x^2(x^2+5)^{-3/2}+5(x^2+5)^{-3/2}-x^2(x^2+5)^{-3/2}\]
\[y''=5(x^2+5)^{-3/2}\]
\[\frac{5}{y^3} = 5y^{-3} = 5((x^2+5)^{1/2})^{-3}\]
\[5y^{-3}=5(x^2+5)^{-3/2}=y''\]

Answer 4

ty guys :)