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calculus 2 question Differential Equation?
A) dx/dy = e^(5 x)/(3 y^2) and y(0)=3
y(x)=?
Can someone please help I really dont understand :/ can you please include the steps too i will choose best answer so you can get 10 pts
Also:
b) find the solution of the differential equation dx/dy = y^2+9 that satisfies the initial condition y(7)=0
y(x)=?
thanks !
1 Answer
- MechEng2030Lv 77 years agoFavorite Answer
You must've mean dy/dx? Not dx/dy?
A.)
dy/dx = e^(5x)/(3y^2)
Separating variables:
3y^2 dy = e^(5x) dx
Integrating both sides:
y^3 = 1/5*e^(5x) + C
Plugging in initial condition:
3^3 = 1/5 + C
C = 27 - 1/5 = 134/5
So y = (1/5*e^(5x) + 134/5)^(1/3)
B.)
dy/dx = y^2 + 9
dy/(y^2 + 9) = dx
Integrating both sides:
1/3*arctan(y/3) = x + C
Plugging in initial condition:
1/3*arctan(0/3) = 7 + C => C = -7
y/3 = tan(3x - 21)
y = 3*tan(3x - 21)
