Question

Solve the initival value problem: y′=7 cos(5x)/(8−3y)y′=7 cos⁡(5x)/(8−3y), y(0)=3y(0)=3. y=y= When solving an ODE, the solution is only valid in some interval. Furthermore, if an initial condition is given, the solution will only be valid in the largest interval in the domain of the solution that is around the xx-value given in the initial condition. In this case, since y(0)=3y(0)=3, then the solution is only valid in the largest interval in the domain of yy around x=0x=0.

Answer 1

The solution to the differential equation

y' = (7cos5x)/(8 - 3y); y(0) = 3

is

16y - 3y² = 70sin5x + 21

Explanation:

y' = (7cos5x)/(8 - 3y)

This can be written as

dy/dx = (7cos5x)/(8 - 3y)

Separate the variables

(8 - 3y)dy = (7cos5x)dx

Integrate both sides

8y - (3/2)y² = 35sin5x + C

Applying the initial condition y(0) = 3

8(3) - (3/2)(3)² = 35sin(5(0)) + C

24 - (27/2) = 0 + C

C = 21/2

Therefore,

8y - (3/2)y² = 35sin5x + 21/2

Or

16y - 3y² = 70sin5x + 21