Solution of $y''-y'=xe^x$ using Method of Undetermined Coefficients
The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.
My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:
$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$
This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.
differential-equations
asked Jun 28 '17 at 4:18
Matt Chu
112
add a comment |
The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.
My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:
$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$
This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.
differential-equations
asked Jun 28 '17 at 4:18
Matt Chu
112
3
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
5
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19
add a comment |
The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.
My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:
$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$
This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.
differential-equations
asked Jun 28 '17 at 4:18
Matt Chu
112
The problem gives $y''-y'=xe^x$ with conditions $y(0)=2$ and $y'left(0right)=1$.
My issue is when calculating the particular solution of the equation. I assumed that it should take the form $y_p=(Ax+B)e^x$. After substituting into the 2nd ODE, it becomes:
$(2A+Ax+B)e^x-(A+Ax+B)e^x=xe^x$
This simplifies to $Ae^x=xe^x$. I'm not sure how to progress. Is $A=0$? This does not seem like the correct way to proceed with solving the problem but I am not sure how to proceed.
differential-equations
differential-equations
asked Jun 28 '17 at 4:18
Matt Chu
112
asked Jun 28 '17 at 4:18
Matt Chu
112
edited Jun 28 '17 at 4:23
asked Jun 28 '17 at 4:18
Matt Chu
112
asked Jun 28 '17 at 4:18
Matt Chu
112
asked Jun 28 '17 at 4:18
Matt Chu
112
112
3
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
5
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19
add a comment |
3
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
5
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19
3
3
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
5
5
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19
add a comment |
2 Answers
2
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oldest
votes
If I may suggest, proceed by step.
Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
add a comment |
For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.
answered Nov 28 '18 at 8:37
AwWSaM
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
If I may suggest, proceed by step.
Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
add a comment |
If I may suggest, proceed by step.
Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
add a comment |
If I may suggest, proceed by step.
Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
If I may suggest, proceed by step.
Start using $y'=u$ to have $$y''-y'=xe^ximplies u'-u=xe^x$$ Now let $u=v e^x$ to get $$v' e^x=x e^x implies v'=ximplies v=frac 12 x^2+c_1$$ Here, you already see that the solution will contain an $x^2$ term somewhere.
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
answered Jun 28 '17 at 4:59
Claude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.
answered Nov 28 '18 at 8:37
AwWSaM
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
add a comment |
For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.
answered Nov 28 '18 at 8:37
AwWSaM
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
add a comment |
For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.
answered Nov 28 '18 at 8:37
AwWSaM
1
For any Problem First try taking the RHS as y, In this case y=xe^x. Substitute in the LHS, If you get the answer as 0 then your guess is simply not compatible, It should be yp=(Ax+B)xe^x. To be exact you should multiply your guess by x until you find a different league of terms than in the LHS.
answered Nov 28 '18 at 8:37
AwWSaM
1
answered Nov 28 '18 at 8:37
AwWSaM
1
answered Nov 28 '18 at 8:37
AwWSaM
1
answered Nov 28 '18 at 8:37
AwWSaM
1
1
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
add a comment |
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
– José Carlos Santos
Nov 28 '18 at 8:58
add a comment |
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3
Try $(Ax^2+Bx+C)e^x$.
– Lord Shark the Unknown
Jun 28 '17 at 4:20
@LordSharktheUnknown Why wouldn't it take the form $(Ax+B)e^x$ though?
– Matt Chu
Jun 28 '17 at 4:21
5
I thought you'd just proved why not.
– Lord Shark the Unknown
Jun 28 '17 at 4:26
It doesn't take the form $(Ax + B)e^x because of the second derivative.
– Retired account
Jun 28 '17 at 4:43
The reason you don't have $(Ax+B)e^x$ is that $e^x$ is already a solution to the homogeneous equation $y''-y'=0$, so you need to find another term to use instead.
– Carmeister
Jun 28 '17 at 5:19