Can someone help me out with this questions? I am completely lost.
A) If y=ln{x+(x^2-a^2)}, evaluate dy/dx, expressing your answer in its simplest form.
Show that (x^2 - a^2) d^(2)y/dx^2 +x(dy/dx)=0
Can someone help me out with this questions? I am completely lost.
A) If y=ln{x+(x^2-a^2)}, evaluate dy/dx, expressing your answer in its simplest form.
Show that (x^2 - a^2) d^(2)y/dx^2 +x(dy/dx)=0
$\displaystyle y = \ln\left[x + (x^2-a^2)\right]$
$\displaystyle \frac{dy}{dx} = \frac{2x+1}{x+(x^2-a^2)}$
$\displaystyle \frac{d^2y}{dx^2} = \frac{2[x+(x^2-a^2)] - (2x+1)^2}{[x + (x^2-a^2)]^2} = -\frac{2x^2+2x+2a^2+1}{[x+(x^2-a^2)]^2}$
$\displaystyle x \cdot \frac{dy}{dx} = \frac{x(2x+1)}{x+(x^2-a^2)}$
$\displaystyle (x^2-a^2) \cdot \frac{d^2y}{dx^2} = \frac{(a^2-x^2)(2x^2+2x+2a^2+1)}{[x+(x^2-a^2)]^2}$
... now, see if you can get the last two expressions to add up to 0.