y''=(y')^2-y

Question

solve differential equations of higher orders that allow the lowering of powers ,

Cauchy condition
y (1) = - 1/4
n (1) = 1/2

 

Answer 1

Solve the autonomous equation (d^2 y(x))/(dx^2) = ((dy(x))/(dx))^2 - y(x):

Treating y as the independent variable, let v(y) = (dy(x))/(dx) which gives (d^2 y(x))/(dx^2) = d/(dx) ((dy(x))/(dx)) = (dv(y))/(dx) = (dv(y))/(dy) (dy)/(dx) = v(y) (dv(y))/(dy):

(dv(y))/(dy) v(y) = -y + v(y)^2

Subtract v(y)^2 from both sides:

(dv(y))/(dy) v(y) - v(y)^2 = -y

Multiply both sides by 2:

2 (dv(y))/(dy) v(y) - 2 v(y)^2 = -2 y

Let u(y) = v(y)^2, which gives (du(y))/(dy) = 2 v(y) (dv(y))/(dy):

(du(y))/(dy) - 2 u(y) = -2 y

Let μ(y) = e^( integral-2 dy) = e^(-2 y).

Multiply both sides by μ(y):

e^(-2 y) (du(y))/(dy) - (2 e^(-2 y)) u(y) = -2 e^(-2 y) y

Substitute -2 e^(-2 y) = d/(dy) (e^(-2 y)):

e^(-2 y) (du(y))/(dy) + d/(dy) (e^(-2 y)) u(y) = -2 e^(-2 y) y

Apply the reverse product rule f (dg)/(dy) + g (df)/(dy) = d/(dy) (f g) to the left-hand side:

d/(dy) (e^(-2 y) u(y)) = -2 e^(-2 y) y

Integrate both sides with respect to y:

integral d/(dy) (e^(-2 y) u(y)) dy = integral-2 e^(-2 y) y dy

Evaluate the integrals:

e^(-2 y) u(y) = -2 e^(-2 y) (-y/2 - 1/4) + c_1, where c_1 is an arbitrary constant.

Divide both sides by μ(y) = e^(-2 y):

u(y) = y + c_1 e^(2 y) + 1/2

Solve for v(y) in u(y) = v(y)^2:

v(y) = -sqrt(y + c_1 e^(2 y) + 1/2) or v(y) = sqrt(y + c_1 e^(2 y) + 1/2)

Substitute back for (dy(x))/(dx) = v(y):

(dy(x))/(dx) = -sqrt(c_1 e^(2 y(x)) + y(x) + 1/2) or (dy(x))/(dx) = sqrt(c_1 e^(2 y(x)) + y(x) + 1/2)

For (dy(x))/(dx) = -sqrt(e^(2 y(x)) c_1 + y(x) + 1/2):

Divide both sides by sqrt(e^(2 y(x)) c_1 + y(x) + 1/2):

((dy(x))/(dx))/sqrt(c_1 e^(2 y(x)) + y(x) + 1/2) = -1

Integrate both sides:

integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = integral_(c_2)^x-1 du

Evaluate the integrals:

integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = -x + c_2, where c_2 is an arbitrary constant.

For (dy(x))/(dx) = sqrt(e^(2 y(x)) c_1 + y(x) + 1/2):

Divide both sides by sqrt(e^(2 y(x)) c_1 + y(x) + 1/2):

((dy(x))/(dx))/sqrt(c_1 e^(2 y(x)) + y(x) + 1/2) = 1

Integrate both sides:

integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = integral_(c_2)^x 1du

Evaluate the integrals:

integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = x + c_2, where c_2 is an arbitrary constant.

Collect solutions:

Answer: |

| integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = -x + c_2 or integral_1^y(x) 1/sqrt(u + e^(2 u) c_1 + 1/2) du = x + c_2