Solving Differential Equations?

2015-01-19 5:09 am
The differential equation

exp(y)dy/dx = (4x+4)/(-1siny+9cosy)

has an implicit general solution of the form F(x,y)= K

In fact, because the differential equation is separable, we can define the solution curve implicitly by a function in the form F(x,y)= G(x)+H(y)=K

Find such a solution and then give the related functions requested.


How I did it:
integral(e^y(-siny+9cosy))dy= integral(4x+4)dx

RHS = 2x(x+2)+C
LHS = e^y(4cosy+5siny)

F(x,y)= e^y(4cosy+5siny)-2x(x+2) = K

However, my answer is not correct.

回答 (2)

✔ 最佳答案
e^(y) * dy/dx = (4x + 4) / (9cos(y) - sin(y))
e^(y) * (9cos(y) - sin(y)) * dy = 4 * (x + 1) * dx

9 * cos(y) - sin(y) =>
sqrt(9^2 + (1)^2) * ((9/sqrt(82)) * cos(y) - sin(y) * (1/sqrt(82))) =>
sqrt(82) * ((9/sqrt(82)) * cos(y) - (1/sqrt(82)) * sin(y))

sin(t) = 9/sqrt(82)
cos(t) = 1/sqrt(82)

sqrt(82) * (sin(t)cos(y) - sin(y)cos(t)) =>
sqrt(82) * sin(t - y) =>
-sqrt(82) * sin(y - t)

e^(y) * (9cos(y) - sin(y)) * dy = 4 * (x + 1) * dx
e^(y) * -sqrt(82) * sin(y - t) * dy = 4 * (x + 1) * dx
e^(y) * sin(y - t) * dy = (-4/sqrt(82)) * (x + 1) * dx
e^(y) * sin(y - t) * dy = (-4 * sqrt(82) / 82) * (x + 1) * dx
e^(y) * sin(y - t) * dy = (-2 * sqrt(82) / 41) * (x + 1) * dx

t is just some unknown constant for now

u = e^(y)
du = e^(y) * dy
dv = sin(y - t) * dy
v = -cos(y - t)

int(e^(y) * sin(y - t) * dy) = -e^(y) * cos(y - t) + int(e^(y) * cos(y - t) * dy)

u = e^(y)
du = e^(y) * dy
dv = cos(y - t) * dy
v = sin(y - t)

int(e^(y) * sin(y - t) * dy) = -e^(y) * cos(y - t) + int(e^(y) * cos(y - t) * dy)
int(e^(y) * sin(y - t) * dy) = -e^(y) * cos(y - t) + e^(y) * sin(y - t) - int(e^(y) * sin(y - t) * dt)
2 * int(e^(y) * sin(y - t) * dy) = e^(y) * (sin(y - t) - cos(y - t))
int(e^(y) * sin(y - t) * dy) = (1/2) * e^(y) * (sin(y - t) - cos(y - t))

e^(y) * sin(y - t) * dy = (-2 * sqrt(82) / 41) * (x + 1) * dx
(1/2) * e^(y) * (sin(y - t) - cos(y - t)) = (-2 * sqrt(82) / 41) * ((1/2) * x^2 + x) + C
e^(y) * (sin(y - t) - cos(y - t)) = 2 * (-2 * sqrt(82) / 41) * (1/2) * (x^2 + 2x) + C
e^(y) * (sin(y - t) - cos(y - t)) = (-2 * sqrt(82) / 41) * (x^2 + 2x) + C

sqrt(2) * (sin(y - t) * sqrt(2)/2 - sqrt(2)/2 * cos(y - t)) =>
sqrt(2) * (sin(y - t) * cos(pi/4) - sin(pi/4) * cos(y - t)) =>
sqrt(2) * sin(y - t - pi/4)

e^(y) * (sin(y - t) - cos(y - t)) = (-2 * sqrt(82) / 41) * (x^2 + 2x) + C
e^(y) * sqrt(2) * sin(y - t - pi/4) = (-2 * sqrt(82) / 41) * (x^2 + 2x) + C
e^(y) * sin(y - t - pi/4) = (-2 * sqrt(2) * sqrt(41) / (41 * sqrt(2))) * (x^2 + 2x) + C
e^(y) * sin(y - t - pi/4) = (-2 * sqrt(41) / 41) * (x^2 + 2x) + C

sin(t) = 9/sqrt(82)
cos(t) = 1/sqrt(82)

t is in Q1

t = arcsin(9 / sqrt(82))

e^(y) * sin(y - t - pi/4) = (-2 * sqrt(41) / 41) * (x^2 + 2x) + C
e^(y) * sin(y - arcsin(9 / sqrt(82)) - (pi/4)) + (2 * sqrt(41) / 41) * (x^2 + 2x) = K

G(x) = (2 * sqrt(41) / 41) * (x^2 + 2x)
H(y) = e^(y) * sin(y - arcsin(9 / sqrt(82)) - (pi/4))


We can expand that sin(k) out, if you'd like

sin(y - arcsin(9 / sqrt(82)) - pi/4) =>
sin(y - t) * cos(pi/4) - sin(pi/4) * cos(y - t) =>
(sqrt(2)/2) * (sin(y - t) - cos(y - t)) =>
(sqrt(2)/2) * (sin(y)cos(t) - sin(t)cos(y) - cos(y)cos(t) - sin(t)sin(y)) =>
(sqrt(2)/2) * (sin(y) * sqrt(1 - sin(t)^2) - sin(t) * cos(y) - cos(y) * sqrt(1 - sin(t)^2) - sin(t) * sin(y)) =>
(sqrt(2)/2) * (sin(y) * sqrt(1 - (9/sqrt(82))^2) - (9/sqrt(82)) * cos(y) - cos(y) * sqrt(1 - (9/sqrt(82))^2) - (9/sqrt(82)) * sin(y)) =>
(sqrt(2)/2) * (sin(y) * sqrt(1/82) - (9/sqrt(82)) * cos(y) - cos(y) * sqrt(1/82) - (9/sqrt(82)) * sin(y)) =>
(sqrt(2)/2) * (1/sqrt(82)) * (sin(y) - 9cos(y) - cos(y) - 9sin(y)) =>
(1/sqrt(41)) * (-8sin(y) - 10cos(y)) =>
(-2 * sqrt(41) / 41) * (4sin(y) + 5cos(y))

e^(y) * sin(y - arcsin(9 / sqrt(82)) - (pi/4)) + (2 * sqrt(41) / 41) * (x^2 + 2x) = K
e^(y) * (-2sqrt(41) / 41) * (4sin(y) + 5cos(y)) + (2sqrt(41) / 41) * (x^2 + 2x) =>
(2 * sqrt(41) / 41) * (x^2 + 2x - e^(y) * (4sin(y) + 5cos(y))
2015-01-20 2:19 am
The answer is: 2x(x+2)-e^y(4siny+5cosy)


收錄日期: 2021-04-15 18:04:25
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