Ask Question
5 March, 10:07

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.

+1
Answers (1)
  1. 5 March, 12:50
    0
    The solution to the differential equation

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

    is

    16y - 3y² = 70sin5x + 21

    Step-by-step 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
Know the Answer?
Not Sure About the Answer?
Get an answer to your 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 ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers