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

$\frac{dy}{dx}=\frac{ycos(x)}{1+y^2}$

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

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

Now, we can do the integration

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

Then the equation is

$ln(y)+\frac{1}{2}y^2=sin(x)+C$

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

That means at x = 0, y = 1

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

The equation is

$\frac{1}{2}y^2+ln(y)=sin(x)+\frac{1}{2}$