solve-Differential-equation-x-2-d-2-y-dx-2-4x-dy-dx-3y-x-3-2x-5-

Question Number 76964 by peter frank last updated on 01/Jan/20

s o l v e D i f f e r e n t i a l e q u a t i o n

x^{2} \frac{d^{2} y}{{d x}^{2}} - 4 x \frac{d y}{d x} + 3 y = x^{3} + 2 x + 5

Answered by mind is power last updated on 02/Jan/20

x^{2} \frac{d^{2} y}{{dx}^{2}} - 4 x \frac{dy}{dx} + 3 y = 0 . . E

theorie of linear algebra applied in equation

we know that solution of linear Differtionsl equation

forme vectoriel subspace of \dim 2

let y = x^{m}

\Rightarrow ((m (m - 1) - 4 m + 3) = 0

\Rightarrow m^{2} - 5 m + 3 = 0

m_{1} = \frac{5 - \sqrt{13}}{2}, m_{2} = \frac{5 + \sqrt{13}}{2}

y_{1} = x^{\frac{5 - \sqrt{13}}{2}}, y_{2} = x^{\frac{5 + \sqrt{13}}{2}} are independente \Rightarrow

(y_{1}, y_{2}) Bases of solutions of E

General solution of E is, \forall y \in E \exists! (a, b) \in R^{2} ∣ y = {ax}^{\frac{5 - \sqrt{13}}{2}} + {bx}^{\frac{5 + \sqrt{13}}{2}}

particulare solution

P (x) = {ax}^{3} + {bx}^{2} + cx + d

\Rightarrow (6 a - 12 a + 3 a) = 1

(2 b - 8 b + 3 b) = 0

(- 4 c + 3 c) = 2

3 d = 5

a = - \frac{1}{3}

b = 0

c = - 2

d = \frac{5}{3}

p (x) = \frac{- x^{3}}{3} - 2 x + \frac{5}{3}

Y = {ax}^{\frac{5 - \sqrt{13}}{2}} + {bx}^{\frac{5 + \sqrt{13}}{2}} \frac{- x^{3} - 6 x + 5}{3}, (a, b) \in R^{2}

Commented by peter frank last updated on 03/Jan/20

t h a n k y o u

Commented by peter frank last updated on 03/Jan/20

w h e r e t h i s 6 a - 12 a + 3 a c a m e f r p m