Solve the IVP with Laplace Transform:

$\{\begin{array}{l}y"+4{y}^{\prime}+4y=(3+t){e}^{-2t}\\ y(0)=2\\ {y}^{\prime}(0)=5\end{array}$

tinfoQ
2020-12-21
Answered

Solve the IVP with Laplace Transform:

$\{\begin{array}{l}y"+4{y}^{\prime}+4y=(3+t){e}^{-2t}\\ y(0)=2\\ {y}^{\prime}(0)=5\end{array}$

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Sadie Eaton

Answered 2020-12-22
Author has **104** answers

asked 2020-11-10

Find the sokaton 02 the ger Initial value provsem. $y\prime -y=8t2,y(0)=1$

asked 2022-06-23

I'm supposed to find the solution of

$(xy-1)\frac{dy}{dx}+{y}^{2}=0$

The solution in the book manipulates it to

$\frac{dx}{dy}+\frac{x}{y}=\frac{1}{{y}^{2}}$

and proceeds to use an integrating factor to solve it to

$xy=\mathrm{ln}y+c$

Which makes sense, but I did it by isolating the differential of y, substituting $xy=v$, and simplifying it to

$\frac{1-v}{v}dv=\frac{dx}{x}$

which gives, on solving,

$\mathrm{ln}y+c=\mathrm{ln}{x}^{2}+\frac{y}{x}$

Which isn't the same.

I can't see any flaw in what I did except for dividing variables throughout the equation (which I see everywhere in methods of solving differential equations, without a clue as to why.)

Can anyone help me with where I got it wrong?

EDIT: I'll add some detail on my steps;

$xy=c\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=\frac{-{y}^{2}}{xy-1}\phantom{\rule{0ex}{0ex}}x\frac{dv}{dx}=\frac{{v}^{2}}{1-v}+v=\frac{v}{1-v}\phantom{\rule{0ex}{0ex}}\frac{dx}{x}=\frac{(1-v)dv}{v}$

$(xy-1)\frac{dy}{dx}+{y}^{2}=0$

The solution in the book manipulates it to

$\frac{dx}{dy}+\frac{x}{y}=\frac{1}{{y}^{2}}$

and proceeds to use an integrating factor to solve it to

$xy=\mathrm{ln}y+c$

Which makes sense, but I did it by isolating the differential of y, substituting $xy=v$, and simplifying it to

$\frac{1-v}{v}dv=\frac{dx}{x}$

which gives, on solving,

$\mathrm{ln}y+c=\mathrm{ln}{x}^{2}+\frac{y}{x}$

Which isn't the same.

I can't see any flaw in what I did except for dividing variables throughout the equation (which I see everywhere in methods of solving differential equations, without a clue as to why.)

Can anyone help me with where I got it wrong?

EDIT: I'll add some detail on my steps;

$xy=c\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=\frac{-{y}^{2}}{xy-1}\phantom{\rule{0ex}{0ex}}x\frac{dv}{dx}=\frac{{v}^{2}}{1-v}+v=\frac{v}{1-v}\phantom{\rule{0ex}{0ex}}\frac{dx}{x}=\frac{(1-v)dv}{v}$

asked 2022-06-23

Given a second order differential equation

${y}^{\u2033}+f(y){y}^{\prime}+g(y)=0,$

write an equivalent system of first order equations with transformations

${x}_{1}=y,{x}_{2}={y}^{\prime}+{\int}_{0}^{y}f(s)ds.$

This is what I did:

${x}_{1}^{\prime}={y}^{\prime}={x}_{2}-{\int}_{0}^{y}f(s)ds={x}_{2}-{\int}_{0}^{{x}_{1}}f(s)ds$

${x}_{2}^{\prime}={y}^{\u2033}+f(y)=-f(y){y}^{\prime}-g(y)+f(y)=f(y)(1-{y}^{\prime})-g(y)=f({x}_{1})(1-{x}_{2}+{\int}_{0}^{{x}_{1}}f(s)ds)-g({x}_{1})$

I feel like this answer is wrong though, because I am not sure if I'm doing the standard procedure.

${y}^{\u2033}+f(y){y}^{\prime}+g(y)=0,$

write an equivalent system of first order equations with transformations

${x}_{1}=y,{x}_{2}={y}^{\prime}+{\int}_{0}^{y}f(s)ds.$

This is what I did:

${x}_{1}^{\prime}={y}^{\prime}={x}_{2}-{\int}_{0}^{y}f(s)ds={x}_{2}-{\int}_{0}^{{x}_{1}}f(s)ds$

${x}_{2}^{\prime}={y}^{\u2033}+f(y)=-f(y){y}^{\prime}-g(y)+f(y)=f(y)(1-{y}^{\prime})-g(y)=f({x}_{1})(1-{x}_{2}+{\int}_{0}^{{x}_{1}}f(s)ds)-g({x}_{1})$

I feel like this answer is wrong though, because I am not sure if I'm doing the standard procedure.

asked 2021-09-03

Find the Laplace transform of $\mathrm{sin}ht-\mathrm{cos}ht$

asked 2020-11-30

Solve differential equation
$dy/dx-(cotx)y=si{n}^{3}x$

asked 2022-06-19

Find a one parameter family of solutions of the following first order ordinary differential equation

$(3{x}^{2}+9xy+5{y}^{2})dx-(6{x}^{2}+4xy)dy=0$

Hello. So I am stuck after I find out that they are not exact. Please help.

$(3{x}^{2}+9xy+5{y}^{2})dx-(6{x}^{2}+4xy)dy=0$

Hello. So I am stuck after I find out that they are not exact. Please help.

asked 2022-06-24

I was given the following second-order differential equation,

${y}^{\mathrm{\prime}\mathrm{\prime}}+2{y}^{\mathrm{\prime}}+y=g(t),$

and that the solution is $y(t)=(1+t)(1+{e}^{-t})$. Using the solution I determined that

$g(t)=t+3.$

Following from this I transformed this second-order differential equation into a system of first-order differential equations, which is

$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right)\left(\begin{array}{c}y\\ {y}^{\mathrm{\prime}}\end{array}\right)+\left(\begin{array}{c}0\\ t+3\end{array}\right)$

Now I want to perform a single step with $\mathrm{\Delta}t=1$ starting from t=0 with the Forward Euler method and after that with the Backward Euler method. Firstly with the Forward Euler method I use:

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n},{w}_{n})$

and I compute ${w}_{0}$ as

${w}_{0}=\left(\begin{array}{c}y(0)\\ {y}^{\mathrm{\prime}}(0)\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$

so therefore

${w}_{1}=\left(\begin{array}{c}3\\ 0\end{array}\right)$

Now I want to perform the Backward Euler method.

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n+1},{w}_{n+1})$

so

${w}_{1}=\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right){w}_{1}+\left(\begin{array}{c}0\\ 4\end{array}\right)$

From which i get

${w}_{1}=\frac{1}{4}\left(\begin{array}{c}11\\ 3\end{array}\right)$

y two results seems to be quite differnt and that gets me to believe that I have made a mistake somewhere. Could someone let me know if they believe this to be correct, or why this could be wrong?

${y}^{\mathrm{\prime}\mathrm{\prime}}+2{y}^{\mathrm{\prime}}+y=g(t),$

and that the solution is $y(t)=(1+t)(1+{e}^{-t})$. Using the solution I determined that

$g(t)=t+3.$

Following from this I transformed this second-order differential equation into a system of first-order differential equations, which is

$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right)\left(\begin{array}{c}y\\ {y}^{\mathrm{\prime}}\end{array}\right)+\left(\begin{array}{c}0\\ t+3\end{array}\right)$

Now I want to perform a single step with $\mathrm{\Delta}t=1$ starting from t=0 with the Forward Euler method and after that with the Backward Euler method. Firstly with the Forward Euler method I use:

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n},{w}_{n})$

and I compute ${w}_{0}$ as

${w}_{0}=\left(\begin{array}{c}y(0)\\ {y}^{\mathrm{\prime}}(0)\end{array}\right)=\left(\begin{array}{c}2\\ 1\end{array}\right)$

so therefore

${w}_{1}=\left(\begin{array}{c}3\\ 0\end{array}\right)$

Now I want to perform the Backward Euler method.

${w}_{n+1}={w}_{n}+\mathrm{\Delta}tf({t}_{n+1},{w}_{n+1})$

so

${w}_{1}=\left(\begin{array}{c}2\\ 1\end{array}\right)+\left(\begin{array}{cc}0& 1\\ -1& -2\end{array}\right){w}_{1}+\left(\begin{array}{c}0\\ 4\end{array}\right)$

From which i get

${w}_{1}=\frac{1}{4}\left(\begin{array}{c}11\\ 3\end{array}\right)$

y two results seems to be quite differnt and that gets me to believe that I have made a mistake somewhere. Could someone let me know if they believe this to be correct, or why this could be wrong?