By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$y"-4y\text{\'}=-4t{e}^{2t},{y}_{0}=0,y{\text{\'}}_{0}=1$

$y"+2y\text{\'}+10y=-6{e}^{-t}\mathrm{sin}3t,{y}_{0}=0,y{\text{\'}}_{0}=1$

he298c
2021-02-15
Answered

By using Laplace transforms, solve the following differential equations subject to the given initial conditions.

$y"-4y\text{\'}=-4t{e}^{2t},{y}_{0}=0,y{\text{\'}}_{0}=1$

$y"+2y\text{\'}+10y=-6{e}^{-t}\mathrm{sin}3t,{y}_{0}=0,y{\text{\'}}_{0}=1$

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firmablogF

Answered 2021-02-16
Author has **92** answers

Step 1The given differential equation is,

$y"-4{y}^{\prime}=-4t{e}^{2t}$

where ${y}_{0}=0,{y}_{0}^{\prime}=1$

Take the Laplace Transform on both sides,

$L({y}^{\u2033}-4{y}^{\prime})=L(-4t{e}^{2t})$

$L({y}^{\u2033})-4L({y}^{\prime})=-4L(t{e}^{2t})$

Consider the formulae,

$L({y}^{\u2033})={s}^{2}L(y)-s{y}_{0}-{y}_{0}^{\prime}$

$L({y}^{\prime})=sL(y)-{y}_{0}$

$L(t{e}^{2t})=\frac{1}{(s-2{)}^{2}}$

Step 2

Substitute the values,

$L({y}^{\u2033})-4L({y}^{\prime})=-4L(t{e}^{2t})$

${s}^{2}L(y)-s{y}_{0}-{y}_{0}^{\prime}-4[sL(y)-y0]=-4\cdot \frac{1}{(s-2{)}^{2}}$

${s}^{2}L(y)-s\cdot (0)-(1)-4sL(y)=\frac{(-4)}{(s-2{)}^{2}}$

$L(y)\cdot [{s}^{2}-4s]=1+\frac{(-4)}{(s-2{)}^{2}}$

$L(y)=\frac{{s}^{2}-4s+4-4}{(s-2{)}^{2}({s}^{2}-4s)}$

$L(y)=\frac{1}{(s-2{)}^{2}}$

Step 3

Take the Inverse Laplace Transform both sides,

${L}^{-1}[L(y)]={L}^{-1}\left[\frac{1}{(s-2{)}^{2}}\right]$

$y={L}^{-1}\left[\frac{1}{(s-2{)}^{2}}\right]$

$y=t{e}^{2t}$

Thus, the solution of the given integral is $y=t{e}^{2t}$ii)

The given differential equation is,

${y}^{\u2033}+24{y}^{\prime}+10y=-6{e}^{-t}\mathrm{sin}3t$

$\text{where}{y}_{0}=0,{y}_{0}^{\prime}=1$

Step 4

Take the Laplace Transform on both sides,

$L({y}^{\u2033}+24{y}^{\prime}+10y)=L[-6{e}^{-t}\mathrm{sin}(3t)]$

$L({y}^{\u2033})+24L({y}^{\prime})+10L(y)=-6L[{e}^{-t}\mathrm{sin}(3t)]$

Use the same formulae as mentioned above,

${s}^{2}L(y)-0-1+2[sL(y)-0]+10L(y)=(-6)\cdot \frac{3}{(s+1{)}^{2}+9}$

${s}^{2}L(y)-1+2sL(y)+10L(y)=\frac{-18}{{s}^{2}+2s+1+9}$

$[{s}^{2}+2s+10]L(y)=1-\frac{18}{{s}^{2}+2s+10}$

$L(y)=\frac{{s}^{2}+2s+10-18}{({s}^{2}+2s+10{)}^{2}}$

$L\left(y\right)=\frac{{s}^{2}+2s-8}{({s}^{2}+2s+10{)}^{2}}$

Step 5

Take the Inverse Laplace Transform both sides,

${L}^{-1}\left[L\right(y\left)\right]={L}^{-1}\left[\frac{{s}^{2}+2s-8}{({s}^{2}+2s+10{)}^{2}}\right]$

$y=t{e}^{-t}\mathrm{cos}\left(3t\right)$

Thus, the solution of the given differential equation is $y=t{e}^{-t}\mathrm{cos}\left(3t\right)$

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I'm attempting one of the optional problems in my linear algebra textbook and I am very confused as I have never seen a problem in this form. Question is:

Find the vector form of the equation of the line in ${\mathbb{R}}^{3}$ that passes through the point $P=(-1,1,3)$ and is perpendicular to the plane with general equation $x-3y+2z=5$.

I know the normal vector $\overrightarrow{n}=[1,-3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?

My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\overrightarrow{u}=\overrightarrow{P}Q$ and $\overrightarrow{v}=\overrightarrow{P}R$. Then I would use those to form my vector equation

$\overrightarrow{x}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}$

Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.

I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.

Find the vector form of the equation of the line in ${\mathbb{R}}^{3}$ that passes through the point $P=(-1,1,3)$ and is perpendicular to the plane with general equation $x-3y+2z=5$.

I know the normal vector $\overrightarrow{n}=[1,-3,2]$ from the general equation but i'm not sure how it helps me. Based on the question it would seem this would be parallel to line i'm trying to find?

My thought was to find 2 other points on the plane, say Q and R, by satisfying the general equation and use those points to find the direction vectors $\overrightarrow{u}=\overrightarrow{P}Q$ and $\overrightarrow{v}=\overrightarrow{P}R$. Then I would use those to form my vector equation

$\overrightarrow{x}=\overrightarrow{p}+s\overrightarrow{u}+t\overrightarrow{v}$

Is this the proper way to solve this type of problem? Or would this only work if the line is parallel to the plane? The parallel/perpendicular to the plane concept has me really confused.

I just started this course and i'm struggling, so i'm doing my best to try and understand problems outside of the assigned homework (one which I cannot look up an answer for). If someone could give me the proper advice for how to approach these types of problems, I would be extremely grateful. Thanks.