Solve $d\frac{{d}^{2}y}{d{t}^{2}}\text{}-\text{}8d\frac{dy}{dt}\text{}+\text{}15y=9t{e}^{3t}\text{}with\text{}y(0)=5,\text{}{y}^{\prime}(0)=10$

nagasenaz
2020-12-06
Answered

Solve $d\frac{{d}^{2}y}{d{t}^{2}}\text{}-\text{}8d\frac{dy}{dt}\text{}+\text{}15y=9t{e}^{3t}\text{}with\text{}y(0)=5,\text{}{y}^{\prime}(0)=10$

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toroztatG

Answered 2020-12-07
Author has **98** answers

Given:$\frac{{d}^{2}y}{d{t}^{2}}\text{}-\text{}8\frac{dy}{dt}\text{}+\text{}15y=9t{e}^{3t}$

with$y(0)=5,\text{}{y}^{\prime}(0)=10$

The auxiliary equation is given by${m}^{2}\text{}-\text{}8m\text{}+\text{}15=0$

solving this we get$m=5,\text{}3$

Hence the complimentary function is${y}_{c}={C}_{1}{e}^{-3t}\text{}+\text{}{C}_{2}{e}^{-5t}$

Now The P.I. of given differential equation is

${y}_{p}=\frac{1}{{D}^{2}\text{}-\text{}8D\text{}+\text{}15}9t{x}^{3}t=\frac{1}{(D\text{}-\text{}3)(D\text{}-\text{}5)}9t{e}^{3}t$

$={e}^{3}t\frac{1}{(D\text{}+\text{}3\text{}-\text{}3)(D\text{}+\text{}3\text{}-\text{}3)}9t$

$={e}^{3}t\frac{1}{D(D\text{}-\text{}2)}9t$

$={e}^{3}t\frac{1}{-2D(1\text{}-\text{}\frac{D}{2})}9t$

$=\text{}-\frac{{e}^{3t}}{2}\frac{1}{D}{(1\text{}-\text{}\frac{D}{2})}^{-1}(1\text{}-\text{}D{)}^{-1}9t$

$=\text{}-\frac{{e}^{3t}}{2}\frac{1}{D}(1\text{}+\text{}\frac{D}{2}\text{}+\text{}\cdots )9t$

$=\text{}-\frac{{e}^{-3t}}{2}\frac{1}{D}(9t\text{}+\text{}\frac{9}{2})$

$=\text{}-\frac{{e}^{-3t}}{2}\frac{1}{D}(\frac{9}{2}{t}^{2}+\text{}\frac{9}{2}t)$

$=\text{}-\frac{9}{4}{e}^{-3}t(t\text{}+\text{}{t}^{2})$

Hence the general solution is$y={y}_{c}\text{}+\text{}yp={C}_{1}{v}^{-3t}\text{}+\text{}{C}_{2}{e}^{5t}\text{}-\text{}\frac{9}{2}{e}^{-3}t(t\text{}+\text{}{t}^{2})$

Now$y(0)=0\text{}\Rightarrow \text{}{C}_{1}\text{}+\text{}{e}_{2}=5\cdots $ (1)

Also${y}^{\prime}=3{C}_{1}{e}^{-2t}\text{}+\text{}5{C}_{2}{e}^{5t}\text{}-\text{}\frac{9}{4}[3{e}^{-3t}(1\text{}+\text{}2t)=3{e}^{3t}(t\text{}+\text{}{t}^{2})]$

Hence${y}^{\prime}(0)=10\text{}\Rightarrow \text{}3{C}_{1}\text{}+\text{}5{C}_{2}\text{}-\text{}\frac{9}{4}=10\text{}\Rightarrow \text{}12{C}_{1}\text{}+\text{}20{C}_{2}=49\text{}\cdots $ (2) solving (1) and (2) we get

${C}_{1}=\frac{51}{8},\text{}{C}_{2}=\frac{-11}{8}$

Hence the solution is$y=\frac{51}{8}{e}^{3t}\text{}-\text{}\frac{11}{8}{e}^{5t}\text{}-\text{}\frac{9}{4}{e}^{-3t}(t\text{}+\text{}{t}^{2})$

with

The auxiliary equation is given by

solving this we get

Hence the complimentary function is

Now The P.I. of given differential equation is

Hence the general solution is

Now

Also

Hence

Hence the solution is

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Consider the linear first order non-homogeneous partial differential equation

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

${U}_{x}+y{U}_{y}-y=y{e}^{-x}$

By using the method of characteristics show that its general solution is given by

$u(x,y)=-y{e}^{-x}+{e}^{x}g(y{e}^{-x})$

where g is any differentiable function of its argument

$A=1,B=y,d=-1,f=y{e}^{-x}\phantom{\rule{0ex}{0ex}}{\displaystyle \frac{dy}{dx}}={\displaystyle \frac{b}{a}}=y\phantom{\rule{0ex}{0ex}}dy=ydx\phantom{\rule{0ex}{0ex}}\int {\displaystyle \frac{1}{y}}dy=\int dx\phantom{\rule{0ex}{0ex}}\mathrm{ln}y=x+c\phantom{\rule{0ex}{0ex}}\eta (x,y)=c\phantom{\rule{0ex}{0ex}}\text{let}\xi =x$

Then

$\eta =\mathrm{ln}y-x$

$\xi =x$

those are my characteristics curves,

when I put it into $[a{\xi}_{x}+b{\xi}_{y}]{\omega}_{\xi}+d\omega =F$

I get the equation ${\omega}_{\xi}-\omega ={e}^{\eta}$

Is this all correct?

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