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zeotropojd

Answered question

2021-12-26

Solve the following differential equations for a general solution:

y 3 y^{'} + 2 y = e^{x}

Answer & Explanation

Kayla Kline

Beginner2021-12-27Added 37 answers

Given differential equation:

y 3 y^{'} + 2 y = e^{x}

(1)

y 3 y^{'} + 2 y = 0

(2)
The characteristic equation of (2) is

m^{2} + 3 m + 2 = 0

m = \frac{- 3 \pm \sqrt{9 - 8}}{2}

m = \frac{- 3 \pm 1}{2}

m = \frac{- 2}{2} and - 2

m = - 1 and - 2

therefore general solution of (2)

y (x) = c_{1} e^{- x} + c_{2} e^{- 2 x}

If we assume a particular solution of non homogeneous equation is of the form.

y_{p} (x) = k e^{x}

then

y_{p} = k e^{x}, y_{p} k e^{x}

Putting these values in (1), we have

k e^{x} + 3 (k e^{x}) + 2 (k e^{x}) = e^{x}

\Rightarrow (k + 2 k + 3 k) e^{x} = e^{x}

\Rightarrow (k + 2 k + 3 k) = 1

\Rightarrow 6 k = 1

\Rightarrow k = \frac{1}{6}

therefore

y_{p} = \frac{1}{6} e^{x}

.
Hence: General solution of equation (1) is

y (x) = c_{1} e^{- x} + c_{2} e^{- 2 x} + \frac{1}{6} e^{x}

where

c_{1} and c_{2}

are constant.

David Clayton

Beginner2021-12-28Added 36 answers

y = \exp (k \cdot x)

k^{2} - 3 \cdot k + 2 = 0

k_{1} = 2

k_{2} = 1

y = C 1 \cdot \exp (x) + C 2 \cdot \exp (2 \cdot x)

y = C 1 \cdot \exp (x) + C 2 \cdot \exp (2 \cdot x) - x \cdot \exp (x)

karton

Expert2022-01-09Added 613 answers

Complimentary function
Auxiliary equation $r^{2} + 3 r + 2 = 0 \Rightarrow (r + 2) (r + 1) = 0 \Rightarrow r = - 2, - 1$
$y_{c} = A e^{- x} + B e^{- 2 x}$
Particular Integral
$(D + 1) (D + 2) y = e^{e x}; Let (D + 2) y = z$
$(D + 1) = e^{e x}$
$\frac{d z}{d x} + z = e^{e x}$
$I . F . = e^{\int d x} = e^{x}$
Multiplying by I.F. $e^{x} z = \int e^{x} e^{e x} d x$
Put $e^{x} = t; e^{x} d x = d t$
$e^{x} z = \int e^{t} d t = e^{t} = e^{e x}$
$z = e^{- x} e^{e x}$
$(D + 2) y = e^{- x} e^{e x}$
$\frac{d y}{d x} + 2 y = e^{- x} e^{e x}$
$I . F = e^{\int 2 d x} = e^{2 x}$
Multiplying by I.F. and integrating
$e^{2 x} y = \int e^{2 x} e^{- x} e^{e x} d x = \int e^{x} e^{e x} d x = e^{e x}$ as above.
So $y_{p} = e^{- 2 x} e^{e x}$
The total solution is $y = A e^{- x} + B e^{- 2 x} + e^{- 2 x} e^{e x}$

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