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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

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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 + C₁

the General Solution is:

xy² + x cos x = C

substituting the initial conditions:
x = 1 ; y = pi

1 * (pi)² + 1* cos 1 = C

1 * (3.14159)² + 1* 0.54030230586 = C

1 * 9.8696044010894 + 1* 0.5403023058681= C

9.8696044010894 + 0.5403023058681= C

10.4099067069575 = C

C = 10.4099067069575

the particular solution is:

xy² + x cos x = C

xy² + x cos x = 10.4099067069575

or

xy² + x cos x = (pi)² + cos 1



Final Answer:

xy² + x cos x = 10.4099067069575

or

xy² + x cos x = (pi)² + cos 1





Thanks for writing.

Staff
www.solving-math-problems.com

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