Question
Solve the following initial value problems:
$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$
$(\text{x}^2+\text{y}^2)\text{dx}=2\text{xy dy, y}(1)=0$
$\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}^2+\text{v}^2\text{x}^2}{2\text{xvx}}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2\text{v}}-\text{v}$ $\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2-2\text{v}^2}{2\text{v}}$ $\text{x}\frac{\text{dv}}{\text{dx}}=\frac{1-\text{v}^2}{2\text{v}}$ $\int\frac{2\text{v}}{1-\text{v}^2}=\int\frac{\text{dx}}{\text{x}}$ $\log|1-\text{v}^2|=-\log|\text{x}|+\log|\text{C}|$ $\log|1-\text{v}^2|=\log\Big|\frac{\text{C}}{\text{x}}\Big|$ $\Big|\frac{\text{x}^2-\text{y}^2}{\text{x}^2}\Big|=\Big|\frac{\text{C}}{\text{x}}\Big|$ $|\text{x}^2-\text{y}^2|=|\text{Cx}|\ \dots(\text{i})$ Put y = 0, x = 1 1 - 0 = C C = 1 Put the value of C in equation (i), $|\text{x}^2-\text{y}^2|=|\text{x}|$ $(\text{x}^2-\text{y}^2)^2=\text{x}^2$Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\int\frac{\sin^{-1}\text{x}}{\text{x}^2}\text{dx}$