How do I solve (y') ^2 + (x + 2y) cos (x + y) = (x + 2y + cos (x + y)) y'?

Answer 1

If you treat y' in the equation like you would treat x in the quadratic equation, you will find 2 answers for y':

y' = cos(x+y) is shown to be a correct answer if you directly substitute it into the above equation.

however, this is a tough equation to solve in closed form!

the other solution is y' = x + 2y or y' - 2y = x ( You can also show that this y' also satisfies the equation!) No initial conditions given so assume y(0) = A

Then, this equation is solved as y = Ae^2x + (1/4)e^2x - (1/2)x - (1/4)

Answer 2

(y')² + (x + 2y) cos(x + y) = (x + 2y + cos(x + y)) y'

⇒ (y')² + (x + 2y) cos(x + y) = (x + 2y) y' + cos(x + y) y'

⇒ (y')² - cos(x + y) y' = (x + 2y) y' - (x + 2y) cos(x + y)

⇒ y' [y' - cos(x + y)] = (x + 2y) [y' - cos(x + y)]

⇒ (y' - x - 2y) [y' - cos(x + y)] = 0

y' - x - 2y = 0 ----------(1) or

y' - cos(x + y) = 0 ----------(2)

From (1) we get: y' - 2y = x

Let, the integration factor μ(x) = e^{∫-2 dx} = e^-2x

Multiplying both sides by μ(x):

e^-2x(y') - 2e^-2x(y) = e^-2x(x) --------(3)

Let, e^-2x(y) = z

∴ z' = e^-2x(y') - 2e^-2x(y)

So, equation (3) becomes:

z' = e^-2x(x)

⇒ dz/dx = e^-2x(x)

⇒ z = ∫e^-2x(x) dx + C₁

⇒ z = -(1/4)e^-2x(2x + 1) + C₁ [integrating by parts]

⇒ e^-2x(y) = -(1/4)e^-2x(2x + 1) + C₁

⇒ y = C₁e^2x - x/2 - 1/4

From (2) we get:

y' = cos(x + y) --------(4)

Let, z = x + y

∴ z' = y' + 1

So, equation (4) becomes:

z' - 1= cos(z)

⇒ dz/dx = 1 + cos(z)

⇒ ∫dz/(cos(z) + 1) = x + C₂

⇒ tan(z/2) = x + C₂ [integrating by substituting tan(z/2) with some arbitrary variable u]

⇒ z = 2 tan^-1(x + C₂)

⇒ x + y = 2 tan^-1(x + C₂)

⇒ y = 2 tan^-1(x + C₂) - x

y = C₁e^2x - x/2 - 1/4

or

y = 2 tan^-1(x + C₂) - x