Question

Solve the differential equation dy/dx = ycos(x)/1 + y^2 with the initial condition y(0) = 1.

Answer 1

The given differential equation is

`(dy)/(dx)=(ycos(x))/(1+y^2)`

We will separate all terms of y with dy and all terms of x with dx

`\begin{gathered} (1+y^2)/(y)dy=cos(x)dx \\ ((1)/(y)+(y^2)/(y))dy=cos(x)dx \\ ((1)/(y))dy+ydy=cos(x)dx \end{gathered}`

Now, we can do the integration

`\begin{gathered} \int(1)/(y)dy=ln(y) \\ \int ydy=(y^(1+1))/(1+1)=(y^2)/(2) \end{gathered}`
`\int cos(x)=sin(x)+C`

Then the equation is

`ln(y)+(1)/(2)y^2=sin(x)+C`

To find C we will use y(0) = 1

That means at x = 0, y = 1

`\begin{gathered} ln(1)+(1)/(2)(1)^2=sin(0)+C \\ \\ ln(1)=0 \\ sin(0)=0 \\ \\ 0+(1)/(2)=0+C \\ \\ (1)/(2)=C \end{gathered}`

The equation is

`(1)/(2)y^2+ln(y)=sin(x)+(1)/(2)`