ODE's. Repeated complex roots
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I got a question about his ODE: $(D-3)^3(D^2+25)^2(D-2)y = 7x^2e^{3x} + xsin (5x)+4x^3$. So, in the solutions of the homogeneous eq. appears a complex root ($r = pm5i$) with multiplicity 2. What I need to do with the soluctions?.
ordinary-differential-equations
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
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add a comment |
$begingroup$
I got a question about his ODE: $(D-3)^3(D^2+25)^2(D-2)y = 7x^2e^{3x} + xsin (5x)+4x^3$. So, in the solutions of the homogeneous eq. appears a complex root ($r = pm5i$) with multiplicity 2. What I need to do with the soluctions?.
ordinary-differential-equations
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
$endgroup$
add a comment |
$begingroup$
I got a question about his ODE: $(D-3)^3(D^2+25)^2(D-2)y = 7x^2e^{3x} + xsin (5x)+4x^3$. So, in the solutions of the homogeneous eq. appears a complex root ($r = pm5i$) with multiplicity 2. What I need to do with the soluctions?.
ordinary-differential-equations
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
$endgroup$
I got a question about his ODE: $(D-3)^3(D^2+25)^2(D-2)y = 7x^2e^{3x} + xsin (5x)+4x^3$. So, in the solutions of the homogeneous eq. appears a complex root ($r = pm5i$) with multiplicity 2. What I need to do with the soluctions?.
ordinary-differential-equations
ordinary-differential-equations
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
asked Dec 10 '18 at 4:53
tajiri_numero_1tajiri_numero_1
115
115
add a comment |
add a comment |
1 Answer
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It helps to look at the homogeneous solution first
$$ y_h(x) = (c_1x^2+c_2x+c_3)e^{3x} + (c_4x + c_5)sin (5x) + (c_6x + c_7)cos(5x) + c_8e^{2x} $$
where the 8 constants correspond to the 8th degree characteristic polynomial.
Let's look at each individual term on the non-homogeneous side. Let's say we have $y_h = y_1+y_2+y_3$ and
begin{align}
Ly_1 &= 4x^3 \
Ly_2 &= 7x^2e^{3x} \
Ly_3 &= xsin(5x)
end{align}
where $L = (D-3)^3(D^2+25)^2(D-2)$
The first one is easy enough, since it's not part of the homogeneous solution. It's simply
$$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$
The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. This turns out to be
$$ y_2(x) = (b_1x^2 + b_2 x + b_3)x^3e^{3x} $$
where the quadratic factor corresponds to the $x^2$, and the additional power of $x$ is needed to "offset" the homogeneous solution. Note that this power needs to be at least higher than the highest power of the homogenous solution (i.e. $c_1x^2e^{3x}$), or otherwise equal to the multiplicity of its root.
With that in mind, we can make a guess for the last trigonometric term. Its multiplicity is $2$, therefore
$$ y_3(x) = (d_1 x + d_2)x^2sin (5x) + (d_3 x + d_4)x^2cos(5x) $$
You do need both $sin$ and $cos$ terms, and this particular form works for any linear combination of trig functions in the non-homogeneous function.
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
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1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
add a comment |
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1 Answer
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$begingroup$
It helps to look at the homogeneous solution first
$$ y_h(x) = (c_1x^2+c_2x+c_3)e^{3x} + (c_4x + c_5)sin (5x) + (c_6x + c_7)cos(5x) + c_8e^{2x} $$
where the 8 constants correspond to the 8th degree characteristic polynomial.
Let's look at each individual term on the non-homogeneous side. Let's say we have $y_h = y_1+y_2+y_3$ and
begin{align}
Ly_1 &= 4x^3 \
Ly_2 &= 7x^2e^{3x} \
Ly_3 &= xsin(5x)
end{align}
where $L = (D-3)^3(D^2+25)^2(D-2)$
The first one is easy enough, since it's not part of the homogeneous solution. It's simply
$$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$
The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. This turns out to be
$$ y_2(x) = (b_1x^2 + b_2 x + b_3)x^3e^{3x} $$
where the quadratic factor corresponds to the $x^2$, and the additional power of $x$ is needed to "offset" the homogeneous solution. Note that this power needs to be at least higher than the highest power of the homogenous solution (i.e. $c_1x^2e^{3x}$), or otherwise equal to the multiplicity of its root.
With that in mind, we can make a guess for the last trigonometric term. Its multiplicity is $2$, therefore
$$ y_3(x) = (d_1 x + d_2)x^2sin (5x) + (d_3 x + d_4)x^2cos(5x) $$
You do need both $sin$ and $cos$ terms, and this particular form works for any linear combination of trig functions in the non-homogeneous function.
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
$endgroup$
1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
add a comment |
$begingroup$
It helps to look at the homogeneous solution first
$$ y_h(x) = (c_1x^2+c_2x+c_3)e^{3x} + (c_4x + c_5)sin (5x) + (c_6x + c_7)cos(5x) + c_8e^{2x} $$
where the 8 constants correspond to the 8th degree characteristic polynomial.
Let's look at each individual term on the non-homogeneous side. Let's say we have $y_h = y_1+y_2+y_3$ and
begin{align}
Ly_1 &= 4x^3 \
Ly_2 &= 7x^2e^{3x} \
Ly_3 &= xsin(5x)
end{align}
where $L = (D-3)^3(D^2+25)^2(D-2)$
The first one is easy enough, since it's not part of the homogeneous solution. It's simply
$$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$
The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. This turns out to be
$$ y_2(x) = (b_1x^2 + b_2 x + b_3)x^3e^{3x} $$
where the quadratic factor corresponds to the $x^2$, and the additional power of $x$ is needed to "offset" the homogeneous solution. Note that this power needs to be at least higher than the highest power of the homogenous solution (i.e. $c_1x^2e^{3x}$), or otherwise equal to the multiplicity of its root.
With that in mind, we can make a guess for the last trigonometric term. Its multiplicity is $2$, therefore
$$ y_3(x) = (d_1 x + d_2)x^2sin (5x) + (d_3 x + d_4)x^2cos(5x) $$
You do need both $sin$ and $cos$ terms, and this particular form works for any linear combination of trig functions in the non-homogeneous function.
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
$endgroup$
1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
add a comment |
$begingroup$
It helps to look at the homogeneous solution first
$$ y_h(x) = (c_1x^2+c_2x+c_3)e^{3x} + (c_4x + c_5)sin (5x) + (c_6x + c_7)cos(5x) + c_8e^{2x} $$
where the 8 constants correspond to the 8th degree characteristic polynomial.
Let's look at each individual term on the non-homogeneous side. Let's say we have $y_h = y_1+y_2+y_3$ and
begin{align}
Ly_1 &= 4x^3 \
Ly_2 &= 7x^2e^{3x} \
Ly_3 &= xsin(5x)
end{align}
where $L = (D-3)^3(D^2+25)^2(D-2)$
The first one is easy enough, since it's not part of the homogeneous solution. It's simply
$$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$
The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. This turns out to be
$$ y_2(x) = (b_1x^2 + b_2 x + b_3)x^3e^{3x} $$
where the quadratic factor corresponds to the $x^2$, and the additional power of $x$ is needed to "offset" the homogeneous solution. Note that this power needs to be at least higher than the highest power of the homogenous solution (i.e. $c_1x^2e^{3x}$), or otherwise equal to the multiplicity of its root.
With that in mind, we can make a guess for the last trigonometric term. Its multiplicity is $2$, therefore
$$ y_3(x) = (d_1 x + d_2)x^2sin (5x) + (d_3 x + d_4)x^2cos(5x) $$
You do need both $sin$ and $cos$ terms, and this particular form works for any linear combination of trig functions in the non-homogeneous function.
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
$endgroup$
It helps to look at the homogeneous solution first
$$ y_h(x) = (c_1x^2+c_2x+c_3)e^{3x} + (c_4x + c_5)sin (5x) + (c_6x + c_7)cos(5x) + c_8e^{2x} $$
where the 8 constants correspond to the 8th degree characteristic polynomial.
Let's look at each individual term on the non-homogeneous side. Let's say we have $y_h = y_1+y_2+y_3$ and
begin{align}
Ly_1 &= 4x^3 \
Ly_2 &= 7x^2e^{3x} \
Ly_3 &= xsin(5x)
end{align}
where $L = (D-3)^3(D^2+25)^2(D-2)$
The first one is easy enough, since it's not part of the homogeneous solution. It's simply
$$ y_1(x) = a_1 x^3 + a_2 x^2 + a_3 x + a_4 $$
The second one is a polynomial multiplied with a solution of the homogeneous equation, so we should guess a solution of the same form. This turns out to be
$$ y_2(x) = (b_1x^2 + b_2 x + b_3)x^3e^{3x} $$
where the quadratic factor corresponds to the $x^2$, and the additional power of $x$ is needed to "offset" the homogeneous solution. Note that this power needs to be at least higher than the highest power of the homogenous solution (i.e. $c_1x^2e^{3x}$), or otherwise equal to the multiplicity of its root.
With that in mind, we can make a guess for the last trigonometric term. Its multiplicity is $2$, therefore
$$ y_3(x) = (d_1 x + d_2)x^2sin (5x) + (d_3 x + d_4)x^2cos(5x) $$
You do need both $sin$ and $cos$ terms, and this particular form works for any linear combination of trig functions in the non-homogeneous function.
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
answered Dec 10 '18 at 5:30
DylanDylan
12.8k31026
12.8k31026
1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
add a comment |
1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
1
1
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
$begingroup$
Thank you. You answered my question.
$endgroup$
– tajiri_numero_1
Dec 10 '18 at 5:39
add a comment |
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