Question

xy′ = √(1 − y2 ), y(1) = 0

Answer 1

Answer:

y=sin(ln(x))

Step-by-step explanation:

First, we have to order the terms as follows and express y' as dy / dx:

$x*\frac{dy}{dx} =\sqrt{(1-y^{2} )} \\\frac{x}{dx}=\frac{dy}\sqrt{(1-y^{2} )}}\\\frac{dx}{x}=\frac{dy}{\sqrt{(1-y^{2} )} }$

Then, we have to integrate

$\int{\frac{dx}{x}=\int{\frac{dy}{\sqrt{(1-y^{2} )} }$

with this solution after integration:

$ln(x)+C1=arcsin(y)+C2$

Then, we have to reorder

$arcsin(y)=ln(x)+C$

and applied Sin function on both sides

$sin(arcsin(y))=sin(ln(x)+C)\\y=sin(ln(x)+C)$

To define the value of C, we use the known point y(1)=0 and replace in the equation

$y=sin(ln(x)+C)\\0=sin(ln(1)+C)\\0=sin(0+C)\\0=sin(C)\\C=arcsin(0)\\C=0$

The function that proves that differential equation is

$y=sin(ln(x))$