# particular solution of (cos x - x sin x + y²) dx + 2xy dy = 0

by Rod
(NY)

Differential Equation

• Find the particular solution of:

(cos x - x sin x + y²) dx + 2xy dy = 0

that satisfies the initial condition when x = 1 ; y = pi

### Comments for particular solution of (cos x - x sin x + y²) dx + 2xy dy = 0

 Nov 07, 2012 Differential Equation by: Staff Answer(cos x - x sin x + y²) dx + 2xy dy = 0 `(cos x - x sin x + y²) dx + 2xy dy = 0 ∂ ∂M ----(cos x - x sin x + y²) = 2y = ----- ∂y ∂y ∂ ∂N ----(2xy) = 2y = ----- ∂x ∂x This differential equation (cos x - x sin x + y²) dx + 2xy dy = 0 is exact because ∂M ∂N ------ = ------ = 2y ∂y ∂x f (x, y) = ∫N(x, y) dy = ∫2xy dy = xy² + g(x) ∂fₓ(x, y) = ----[xy² + g(x)] ∂x = y² + g’(x) M (x, y) = (cos x - x sin x + y²) ∴g’(x) = (cos x - x sin x) g(x) = ∫(cos x - x sin x) dx g(x) = x cos x + C₁ ∴ f (x, y) = xy² + x cos x + C₁the General Solution is:xy² + x cos x = Csubstituting the initial conditions: x = 1 ; y = pi1 * (pi)² + 1* cos 1 = C1 * (3.14159)² + 1* 0.54030230586 = C1 * 9.8696044010894 + 1* 0.5403023058681= C9.8696044010894 + 0.5403023058681= C10.4099067069575 = CC = 10.4099067069575the particular solution is:xy² + x cos x = Cxy² + x cos x = 10.4099067069575or xy² + x cos x = (pi)² + cos 1` Final Answer: xy² + x cos x = 10.4099067069575or xy² + x cos x = (pi)² + cos 1 Thanks for writing. Staff www.solving-math-problems.com