If y=(x+x^2-1)^15+(x-x^2-1)^15, then (x^2-1)d^2ydx^2+xdydx is equal to

Question

If $y=\left(x+\sqrt{x^2-1}\right)^{15}+\left(x-\sqrt{x^2-1}\right)^{15}$, then $(x^2-1)\dfrac{d^2y}{dx^2}+x\dfrac{dy}{dx}$ is equal to

Accepted Answer

$\dfrac{dy}{dx}=\dfrac{15y}{\sqrt{x^2-1}}\Rightarrow\sqrt{x^2-1}\,y'=15y$.

Differentiating: $x\dfrac{y'}{\sqrt{x^2-1}}+\sqrt{x^2-1}\,y''=15y'$.

Multiplying by $\sqrt{x^2-1}$: $xy'+(x^2-1)y''=15\sqrt{x^2-1}\,y'=15\cdot15y=225y$.

Solutions · 1