Solve-the-d-e-using-method-of-variation-of-parameter-d-2-y-dx-2-3-dy-dx-2y-sin-e-x-

Question Number 47624 by Umar last updated on 12/Nov/18

Solve the d . e using method of variation

of parameter .

\frac{d^{2} y}{{dx}^{2}} + 3 \frac{dy}{dx} + 2 y = \sin (e^{x})

Answered by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

l e t

t = e^{x}

t = e^{x}

\frac{d t}{d x} = e^{x} = t

\frac{d y}{d x} = \frac{d y}{d t} \times \frac{d t}{d x} = t \frac{d y}{d t}

\frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (t \frac{d y}{d t}) \times \frac{d t}{d x}

= t \times {t \frac{d^{2} y}{{d t}^{2}} + \frac{d y}{d t}} = t^{2} \frac{d^{2} y}{{d t}^{2}} + t \frac{d y}{d t}

n o w

\frac{d^{2} y}{{d x}^{2}} + 3 \frac{d y}{d x} + 2 y = s i n (e^{x})

t^{2} \frac{d^{2} y}{{d t}^{2}} + t \frac{d y}{d t} + 3 t \frac{d y}{d t} + 2 y = s i n (t)

t^{2} \frac{d^{2} y}{{d t}^{2}} + 4 t \frac{d y}{d t} + 2 y = s i n t

n o w

[θ (θ - 1) + 4 θ + 2] y = s i n t θ = \frac{d}{d t}

f o r

C . F

θ^{2} - θ + 4 θ + 2 = 0

(θ + 1) (θ + 2) = 0

θ = - 1, - 2

s o C . F = C_{1} e^{- t} + C_{2} e^{- 2 t} \to C_{1} e^{- e^{x}} + C_{2} e^{- 2 e^{x}}

P . I

y = \frac{1}{θ^{2} + 3 θ + 2} \times s i n t

= \frac{θ^{2} + 2 - 3 θ}{(θ^{2} + 2 + 3 θ) (θ^{2} + 2 - 3 θ)} \times s i n t

= \frac{θ^{2} + 2 - 3 θ}{{(θ^{2} + 2)}^{2} - 9 θ^{2}} s i n t

= \frac{1}{{(- 1^{2} + 2)}^{2} - 9 (- 1^{2})} \times {- s i n t + 2 s i n t - c o s t}

= \frac{1}{1 + 9} \times {s i n t - c o s t}

= \frac{1}{10} \times {s i n (e^{x}) - c o s (e^{x})}

s o c o m p l e t e s o l u t i o n i s

C_{1} e^{- e^{x}} + C_{2} e^{- 2 e^{x}} + \frac{1}{10} {s i n (e^{x}) - c o s (e^{x})}

p l s c h e c k \dots

Commented by Umar last updated on 12/Nov/18

thanks

Commented by tanmay.chaudhury50@gmail.com last updated on 12/Nov/18

m o s t w e l c o m e \dots