Step by step solution to "Find the solution of the following Differential Equations..."

Daniell Phillips

Answered question

2021-12-29

Find the solution of the following Differential Equations

y^{'} - 4 y = 2 x - 4 x^{2}

Answer & Explanation

censoratojk

Beginner2021-12-30Added 46 answers

Find the solution of the following DE
$y^{'} - 4 y = 2 x - 4 x^{2}$
Complete solution $= C F + P I$
CF=y'-4y=0
$(D - 4) y = 0$
$m - 4 = 0$
$m = 4$
$C F = C_{1} e^{4 x}$
$P I = \frac{a}{f (D)} \Rightarrow \frac{2 x - 4 x^{2}}{D - 4}$
$P I = \frac{1}{- 4 + D} (2 x - 4 x^{2})$
$= \frac{1}{- 4 (1 - \frac{D}{4})} (2 x - 4 x^{2}) - 4 + D = - 4 (1 - \frac{D}{4})$
$= - \frac{1}{4} {(1 - \frac{D}{4})}^{- 1} (2 x - 4 x^{2})$
${(1 - D)}^{- 1} = 1 + D + D^{2} + D^{3} + \dots$
${(1 + D)}^{- 1} = 1 - D + D^{2} - D^{3} + \dots$
$D (2 x - 4 x^{2}) = \frac{d}{d x} (2 x - 4 x^{2}) = 2 - 8 x$
$D^{2} (2 x - 4 x^{2}) = \frac{d^{2}}{{d x}^{2}} (2 x - 4 x^{2}) = - 8$
$D^{3} (2 x - 4 x^{2}) = \frac{d^{3}}{{d x}^{3}} (2 x - 4 x^{2}) = 0$
$P I = - \frac{1}{4} (1 + \frac{D}{4} + {(\frac{D}{4})}^{2} + {(\frac{D}{4})}^{3} + \dots) (2 x - 4 x^{2})$

jean2098

Beginner2021-12-31Added 38 answers

y^{'} - 4 y = 2 x - 4 x^{2}

(1)
Clearly, the given differential equation is of the form

y^{'} + p y = a

Thus, differential eq (1) is linear, where

P = - 4 and a = 2 x - 4 x^{2}

Now, integrating factor,

I . F = e^{\int p d x} = e^{\int d x}

\Rightarrow I . F = e^{- 4 x}

The solution of given differential equation (1) is given by,

y (I F) = \int q (I F) d x + c

, c is a constant

\Rightarrow y . e^{- 4 x} = \int (2 x - 4 x^{2}) e^{- 4 x} d x + c

(2)
Now

\int (2 x - 4 x^{2}) e^{- 4 x} d x = \int \frac{e^{u} u (u + 2)}{16} d u

\Rightarrow \int {(2 x - 4 x^{2})}_{e d x}^{- 4 x} \int e^{u} u (u + 2) d u

(3)

\frac{1}{16} \int e^{u} u (u + 2) d u = \frac{1}{16} [(u + 2) (e^{u} u - e^{u}) - \int (e^{u} u - e^{u}) d u]

= \frac{1}{16} [(u + 2) (e^{u} u - e^{u}) - e^{u} u + 2 e^{u}]

= \frac{1}{16} [u^{2} e^{u} - u e^{u} + 2 u e^{u} - 2 e^{u} - u e^{u} + 2 e^{u}]

= \frac{1}{16} u^{2} e^{u}

Hence, from eq (3), we get

\int (2 x - 4 x^{2}) e^{- 4 x} x = \frac{1}{16} u^{2} e^{u}

u = - 4 x

, we get

This is helpful 0 Flag

karton

Expert2022-01-09Added 613 answers

$\begin{matrix} \Rightarrow y . e^{- 4 x} = \int (2 x - 4 x^{2}) e^{- 4 x} d x + c (2) \\ \int (2 x - 4 x^{2}) e^{- 4 x} d x = \int \frac{e^{u} u (u + 2)}{16} d u \\ \Rightarrow \int (2 x - 4 x^{2})_{e d x}^{- 4 x} \int e^{u} u (u + 2) d u (3) \\ \frac{1}{16} \int e^{u} u (u + 2) d u = \frac{1}{16} [(u + 2) (e^{u} u - e^{u}) - \int (e^{u} u - e^{u}) d u] \\ = \frac{1}{16} [(u + 2) (e^{u} u - e^{u}) - e^{u} u + 2 e^{u}] \\ = \frac{1}{16} [u^{2} e^{u} - u e^{u} + 2 u e^{u} - 2 e^{u} - u e^{u} + 2 e^{u}] \\ = \frac{1}{16} u^{2} e^{u} \\ H e n c e, f r o m e q (3), w e g e t \\ i n t (2 x - 4 x^{2}) e^{- 4 x} x = \frac{1}{16} u^{2} e^{u} \\ u = - 4 x, w e g e t \\ \int (2 x - 4 x^{2}) e^{- 4 x} d x = \frac{1}{16} \times (- 4 x)^{2} e^{- 4 x} = x^{2} e^{- 4 x} \\ H e n c e, f r o m e q (2), w e g e t \\ y . e^{- 4 x} = x^{2} e^{- 4 x} + c \\ \Rightarrow y = x^{2} + c e^{4 x} \\ A n s w e r : y = x^{2} + c e^{4 x} \end{matrix}$

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