Solve-the-differential-equation-x-t-2x-t-x-t-1-t-using-the-method-of-variation-of-parameters-

Question Number 107995 by Ar Brandon last updated on 13/Aug/20

Solve the differential equation;

x^{″} (t) + {2 x}^{'} (t) + x (t) = 1 + t

(using the method of variation of parameters)

Answered by Ar Brandon last updated on 13/Aug/20

x_{GH} : r^{2} + 2 r + 1 = 0, {(r + 1)}^{2} = 0, r_{1} = - 1 = r_{2}

\Rightarrow x_{GH} = (At + B) e^{- t} = {Ate}^{- t} + {Be}^{- t}

By varying the parameters,

Let x_{P} = A (t) {te}^{- t} + B (t) e^{- t}

{\begin{cases} A^{'} (t) ({te}^{- t}) + B^{'} (t) (e^{- t}) = 0 & eqn (1) \\ A^{'} (t) {({te}^{- t})}^{'} + B^{'} (t) {(e^{- t})}^{'} = 1 + t & eqn (2) \end{cases}

eqn (2) \Rightarrow A^{'} (t) (e^{- t} - {te}^{- t}) - B^{'} (t) (e^{- t}) = 1 + t

\Rightarrow - [A^{'} (t) ({te}^{- t}) + B^{'} (t) e^{- t}] + A^{'} (t) e^{- t} = 1 + t \Rightarrow A^{'} (t) = e^{t} + {te}^{t}

\Rightarrow A (t) = \int (1 + t) e^{t} dt = (1 + t) e^{t} - e^{t} + k = {te}^{t} + k

A (t) in eqn (1) \Rightarrow B^{'} (t) = - A^{'} (t) t \Rightarrow B (t) = - \int A^{'} (t) tdt

\Rightarrow - B (t) = t \int A^{'} (t) dt - \int {\frac{dt}{dt} \int A^{'} (t) dt} dt

\Rightarrow B (t) = - t ({te}^{t} + k) + \int ({te}^{t} + k) dt

\Rightarrow B (t) = - t ({te}^{t} + k) + [{te}^{t} - e^{t}] + kt + c

\Rightarrow B (t) = - t^{2} e^{t} + {te}^{t} - e^{t} + c

\Rightarrow x_{P} = ({te}^{t} + k) {te}^{- t} + (- t^{2} e^{t} + {te}^{t} - e^{t} + c) e^{- t}

\Rightarrow x_{P} = t^{2} + {kte}^{- t} - t^{2} + t - 1 + {ce}^{- t} = t - 1 + (kt + c) e^{- t}

\Rightarrow x_{G} = (At + B) e^{- t} + t - 1 + (kt + c) e^{- t}

\Rightarrow x_{G} = (α t + β) e^{- t} + t - 1, α = A + k, β = B + c

Answered by Aziztisffola last updated on 13/Aug/20

x^{″} (t) + {2 x}^{'} (t) + x (t) = 0 \Rightarrow r^{2} + 2 r + 1 = 0

{(r + 1)}^{2} = 0 \Rightarrow r = - 1

x_{h} (t) = (α t + β) e^{- t} = α {te}^{- t} + β e^{- t}

x_{p} (t) = u_{1} {te}^{- t} + u_{2} e^{- t}

w = | \begin{matrix} {te}^{- t} & e^{- t} \\ (1 - t) e^{- t} & - e^{- t} \end{matrix} | = - {te}^{- 2 t} - (1 - t) e^{- 2 t}

= (- t - 1 + t) e^{- 2 t} = - e^{- 2 t}

w_{1} = | \begin{matrix} 0 & e^{- t} \\ 1 + t & - e^{- t} \end{matrix} | = - e^{- t} (1 + t)

w_{2} = | \begin{matrix} {te}^{- t} & 0 \\ (1 - t) e^{- t} & 1 + t \end{matrix} | = (1 + t) {te}^{- t}

u_{1} = \int \frac{w_{1}}{w} dt = \int (1 + t) e^{t} dt

u_{2} = \int \frac{w_{2}}{w} dt = - \int (1 + t) {te}^{t} dt

x_{p} (t) = {te}^{- t} \int (1 + t) e^{t} dt - e^{- t} \int (1 + t) {te}^{t} dt

x (t) = x_{h} (t) + x_{p} (t)

Commented by Ar Brandon last updated on 13/Aug/20

Thanks

Answered by mathmax by abdo last updated on 14/Aug/20

laplace method

e \Rightarrow L (y^{^{″}}) + 2 L (y^{^{'}}) + L (y) = L (1 + t) \Rightarrow

t^{2} L (y) - ty (o) - y^{^{'}} (0) + 2 (t L (y) - y (0)) + L (y) = L (1 + t)

\Rightarrow (t^{2} + 2 t + 1) L (y) = ty (o) + 2 y (o) + y^{^{'}} (0) + L (1 + t) \Rightarrow

(t^{2} + 2 t + 1) L (y) = (t + 2) y (o) + y^{^{'}} (0) + L (1 + t) but

L (1 + t) = \int_{0}^{\infty} (1 + x) e^{- tx} dx = \int_{0}^{\infty} e^{- tx} dx + \int_{0}^{\infty} {xe}^{- tx} dx

= {[- \frac{1}{t} e^{- tx}]}_{x = 0}^{\infty} + {[- \frac{x}{t} e^{- tx}]}_{x = 0}^{\infty} + \frac{1}{t} \int_{0}^{\infty} e^{- tx} dx

= \frac{1}{t} + \frac{1}{t} {[- \frac{1}{t} e^{- tx}]}_{x = 0}^{\infty} = \frac{1}{t} + \frac{1}{t^{2}}

e \Rightarrow (t^{2} + 2 t + 1) L (y) = y (0) (t + 2) + y^{^{'}} (0) + \frac{1}{t} + \frac{1}{t^{2}} \Rightarrow

\Rightarrow L (y) = y (0) . \frac{t + 2}{{(t + 1)}^{2}} + y^{^{'}} (0) \frac{1}{{(t + 1)}^{2}} + \frac{1}{t {(t + 1)}^{2}} + \frac{1}{t^{2} {(t + 1)}^{2}} \Rightarrow

y (t) = y (0) L^{- 1} (\frac{t + 2}{{(t + 2)}^{2}}) + y^{^{'}} (0) L^{- 1} (\frac{1}{{(t + 1)}^{2}}) + L^{- 1} (\frac{1}{t {(t + 1)}^{2}})

+ L^{- 1} (\frac{1}{t^{2} {(t + 1)}^{2}}) rest decompisition \dots be continued \dots