Question

Solve the differential equation dy dx equals the quotient of y times the cosine of x and the quantity 1 plus y squared with the initial condition y(0) = 1.

Answer 1

We rearrange the given differential equation to combine all the y terms on the left side of the equation and the x terms on the right side:
     [(1 + y^2) / y] dy = cos(x) dx
     [(1/y) + y] dy = cos(x) dx

We integrate both sides to get
     ln y + (y^2 / 2) = sin(x) + C

Then, we apply the initial condition y = 1 at x = 0 to get the value of C:
     ln (1) + (1^2 / 2) = sin(0) + C
     C = ln (1) + (1^2 / 2) - sin(0)
     C = 1/2

Therefore, our final answer is
     ln (y) + (y^2 / 2) = sin(x) + 1/2