- #1
sami23
- 76
- 1
Solve the Initial Value Problem:
dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1
I need to use substitution and use the formulas:
du/dx = A + B(dy/dx)
dy/dx = (1/B)*(du/dx - A)
Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
therefore A = 3, B = 2
Pluging it into the formula gives me:
(1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)
Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.
How do i make this equation a separable equation?
dy/dx = (3x + 2y)/(3x + 2y + 2) , y(-1) = -1
I need to use substitution and use the formulas:
du/dx = A + B(dy/dx)
dy/dx = (1/B)*(du/dx - A)
Let u = 3x + 2y then du/dx = 3 + 2(dy/dx)
therefore A = 3, B = 2
Pluging it into the formula gives me:
(1/2)*(du/dx - 3) = (3x + 2y)/(3x + 2y + 2)
Now I would need to separate the equation and integrate both sides, replace back what I substituted and solve the IVP where x = -1 and y = -1 to find the constant.
How do i make this equation a separable equation?