Question

Xy′ = √(1 − y2 ), y(1) = 0

Answer 1

y=sin(ln(x))

Explanation:

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

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

Then, we have to integrate

`\int{(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))`