How to find the particular integral of an ODE with an inhomogeneous component formed from the sum of...
$begingroup$
$$y''-3y'+2y=e^{2x}+cos^2x$$
What trial solution is recommended? The standard trial solution would be (unless I'm mistaken):
$$y=ke^{2x}+(asin x+bcos x)(psin x+qcos x)$$
However, the values of $k, a, b, p$ and $q$ would take forever to work out by hand, so there must be an easier method, but I am at a loss on what to do.
I would appreciate any alternative methods for finding the particular integral in this case.
ordinary-differential-equations
edited Jan 2 at 19:52
A.Γ.
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
$endgroup$
add a comment |
$begingroup$
$$y''-3y'+2y=e^{2x}+cos^2x$$
What trial solution is recommended? The standard trial solution would be (unless I'm mistaken):
$$y=ke^{2x}+(asin x+bcos x)(psin x+qcos x)$$
However, the values of $k, a, b, p$ and $q$ would take forever to work out by hand, so there must be an easier method, but I am at a loss on what to do.
I would appreciate any alternative methods for finding the particular integral in this case.
ordinary-differential-equations
edited Jan 2 at 19:52
A.Γ.
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
$endgroup$
add a comment |
$begingroup$
$$y''-3y'+2y=e^{2x}+cos^2x$$
What trial solution is recommended? The standard trial solution would be (unless I'm mistaken):
$$y=ke^{2x}+(asin x+bcos x)(psin x+qcos x)$$
However, the values of $k, a, b, p$ and $q$ would take forever to work out by hand, so there must be an easier method, but I am at a loss on what to do.
I would appreciate any alternative methods for finding the particular integral in this case.
ordinary-differential-equations
edited Jan 2 at 19:52
A.Γ.
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
$endgroup$
$$y''-3y'+2y=e^{2x}+cos^2x$$
What trial solution is recommended? The standard trial solution would be (unless I'm mistaken):
$$y=ke^{2x}+(asin x+bcos x)(psin x+qcos x)$$
However, the values of $k, a, b, p$ and $q$ would take forever to work out by hand, so there must be an easier method, but I am at a loss on what to do.
I would appreciate any alternative methods for finding the particular integral in this case.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 2 at 19:52
A.Γ.
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
edited Jan 2 at 19:52
A.Γ.
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
edited Jan 2 at 19:52
A.Γ.
22.8k32656
edited Jan 2 at 19:52
A.Γ.
22.8k32656
edited Jan 2 at 19:52
A.Γ.
22.8k32656
22.8k32656
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
asked Jan 2 at 19:38
Pancake_SenpaiPancake_Senpai
25116
25116
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The standard method is to split the RHS as
$$
e^{2x}+cos^2x=e^{2x}+frac{1}{2}+frac{1}{2}cos(2x)
$$
and look for a particular integral as the sum
$$
y_1(x)+y_2(x)+y_3(x)
$$
where $y_k$ are particular integrals for each term as RHS separately, i.e. $y_1''-3y_1'+2y_1=e^{2x}$ etc
P.S. Since $e^{2x}$ is a solution to the homogeneous equation, it is not possible to find $y_1$ as $k
e^{2x}$, one would need $k(x)e^{2x}$.
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
Since the complementary function is of the form $$Ae^x+Be^{2x}$$ this is a "failure case", so you can't use $lambda e^{2x}$ as part of the particular integral.
The PI to use is $$lambda xe^{2x}+mu+alphacos2x+betasin2x$$
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
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
$begingroup$
The standard method is to split the RHS as
$$
e^{2x}+cos^2x=e^{2x}+frac{1}{2}+frac{1}{2}cos(2x)
$$
and look for a particular integral as the sum
$$
y_1(x)+y_2(x)+y_3(x)
$$
where $y_k$ are particular integrals for each term as RHS separately, i.e. $y_1''-3y_1'+2y_1=e^{2x}$ etc
P.S. Since $e^{2x}$ is a solution to the homogeneous equation, it is not possible to find $y_1$ as $k
e^{2x}$, one would need $k(x)e^{2x}$.
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
The standard method is to split the RHS as
$$
e^{2x}+cos^2x=e^{2x}+frac{1}{2}+frac{1}{2}cos(2x)
$$
and look for a particular integral as the sum
$$
y_1(x)+y_2(x)+y_3(x)
$$
where $y_k$ are particular integrals for each term as RHS separately, i.e. $y_1''-3y_1'+2y_1=e^{2x}$ etc
P.S. Since $e^{2x}$ is a solution to the homogeneous equation, it is not possible to find $y_1$ as $k
e^{2x}$, one would need $k(x)e^{2x}$.
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
add a comment |
$begingroup$
The standard method is to split the RHS as
$$
e^{2x}+cos^2x=e^{2x}+frac{1}{2}+frac{1}{2}cos(2x)
$$
and look for a particular integral as the sum
$$
y_1(x)+y_2(x)+y_3(x)
$$
where $y_k$ are particular integrals for each term as RHS separately, i.e. $y_1''-3y_1'+2y_1=e^{2x}$ etc
P.S. Since $e^{2x}$ is a solution to the homogeneous equation, it is not possible to find $y_1$ as $k
e^{2x}$, one would need $k(x)e^{2x}$.
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
$endgroup$
The standard method is to split the RHS as
$$
e^{2x}+cos^2x=e^{2x}+frac{1}{2}+frac{1}{2}cos(2x)
$$
and look for a particular integral as the sum
$$
y_1(x)+y_2(x)+y_3(x)
$$
where $y_k$ are particular integrals for each term as RHS separately, i.e. $y_1''-3y_1'+2y_1=e^{2x}$ etc
P.S. Since $e^{2x}$ is a solution to the homogeneous equation, it is not possible to find $y_1$ as $k
e^{2x}$, one would need $k(x)e^{2x}$.
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
answered Jan 2 at 19:43
A.Γ.A.Γ.
22.8k32656
22.8k32656
add a comment |
add a comment |
$begingroup$
Since the complementary function is of the form $$Ae^x+Be^{2x}$$ this is a "failure case", so you can't use $lambda e^{2x}$ as part of the particular integral.
The PI to use is $$lambda xe^{2x}+mu+alphacos2x+betasin2x$$
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
Since the complementary function is of the form $$Ae^x+Be^{2x}$$ this is a "failure case", so you can't use $lambda e^{2x}$ as part of the particular integral.
The PI to use is $$lambda xe^{2x}+mu+alphacos2x+betasin2x$$
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
add a comment |
$begingroup$
Since the complementary function is of the form $$Ae^x+Be^{2x}$$ this is a "failure case", so you can't use $lambda e^{2x}$ as part of the particular integral.
The PI to use is $$lambda xe^{2x}+mu+alphacos2x+betasin2x$$
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
$endgroup$
Since the complementary function is of the form $$Ae^x+Be^{2x}$$ this is a "failure case", so you can't use $lambda e^{2x}$ as part of the particular integral.
The PI to use is $$lambda xe^{2x}+mu+alphacos2x+betasin2x$$
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
answered Jan 2 at 20:07
David QuinnDavid Quinn
24.1k21141
24.1k21141
add a comment |
add a comment |
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