Mixing
Particular integral of $x^2 y''-xy'+4y=cos(ln{x})+xcdot sin(ln{x})$
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I have found the complementary Function, but can't understand how to find the particular integral because with UC method. Please someone help me.
Transcript of image:
Q: $x^2 y'' - xy' + 4y = cos(ln x) + xsin(ln x)$ (labelled equation 1).
Let $x = e^z Rightarrow z = log x$
and $xy' = Delta y$, $x^2y'' = Delta(Delta-1)y$ where $Delta = frac{mathrm{d}}{mathrm{d}z}$
so, equation (1) becomes,
$Delta(Delta - 1)y - Delta y + 4y = cos z + e^zsin z$
$(Delta^2 - 2Delta + 4) y = cos z + e^z sin z$ (labelled equation 2)
Auxilliary equation of equation (2) is
$m^2 - 2m + 4 = 0$
$Rightarrow m = frac{2pm sqrt{4 - 16}}{2}$
$m = 1 pm sqrt{3}i$
Therefore Complementary function, $y_c = e^z c_a cos sqrt{3}z + e^zc_2sinsqrt{3}z$.
ordinary-differential-equations
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
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add a comment |
$begingroup$
I have found the complementary Function, but can't understand how to find the particular integral because with UC method. Please someone help me.
Transcript of image:
Q: $x^2 y'' - xy' + 4y = cos(ln x) + xsin(ln x)$ (labelled equation 1).
Let $x = e^z Rightarrow z = log x$
and $xy' = Delta y$, $x^2y'' = Delta(Delta-1)y$ where $Delta = frac{mathrm{d}}{mathrm{d}z}$
so, equation (1) becomes,
$Delta(Delta - 1)y - Delta y + 4y = cos z + e^zsin z$
$(Delta^2 - 2Delta + 4) y = cos z + e^z sin z$ (labelled equation 2)
Auxilliary equation of equation (2) is
$m^2 - 2m + 4 = 0$
$Rightarrow m = frac{2pm sqrt{4 - 16}}{2}$
$m = 1 pm sqrt{3}i$
Therefore Complementary function, $y_c = e^z c_a cos sqrt{3}z + e^zc_2sinsqrt{3}z$.
ordinary-differential-equations
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
$endgroup$
1
$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36
add a comment |
$begingroup$
I have found the complementary Function, but can't understand how to find the particular integral because with UC method. Please someone help me.
Transcript of image:
Q: $x^2 y'' - xy' + 4y = cos(ln x) + xsin(ln x)$ (labelled equation 1).
Let $x = e^z Rightarrow z = log x$
and $xy' = Delta y$, $x^2y'' = Delta(Delta-1)y$ where $Delta = frac{mathrm{d}}{mathrm{d}z}$
so, equation (1) becomes,
$Delta(Delta - 1)y - Delta y + 4y = cos z + e^zsin z$
$(Delta^2 - 2Delta + 4) y = cos z + e^z sin z$ (labelled equation 2)
Auxilliary equation of equation (2) is
$m^2 - 2m + 4 = 0$
$Rightarrow m = frac{2pm sqrt{4 - 16}}{2}$
$m = 1 pm sqrt{3}i$
Therefore Complementary function, $y_c = e^z c_a cos sqrt{3}z + e^zc_2sinsqrt{3}z$.
ordinary-differential-equations
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
$endgroup$
I have found the complementary Function, but can't understand how to find the particular integral because with UC method. Please someone help me.
Transcript of image:
Q: $x^2 y'' - xy' + 4y = cos(ln x) + xsin(ln x)$ (labelled equation 1).
Let $x = e^z Rightarrow z = log x$
and $xy' = Delta y$, $x^2y'' = Delta(Delta-1)y$ where $Delta = frac{mathrm{d}}{mathrm{d}z}$
so, equation (1) becomes,
$Delta(Delta - 1)y - Delta y + 4y = cos z + e^zsin z$
$(Delta^2 - 2Delta + 4) y = cos z + e^z sin z$ (labelled equation 2)
Auxilliary equation of equation (2) is
$m^2 - 2m + 4 = 0$
$Rightarrow m = frac{2pm sqrt{4 - 16}}{2}$
$m = 1 pm sqrt{3}i$
Therefore Complementary function, $y_c = e^z c_a cos sqrt{3}z + e^zc_2sinsqrt{3}z$.
ordinary-differential-equations
ordinary-differential-equations
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
edited Jan 17 at 17:39
Oscar Lanzi
12.8k12136
12.8k12136
asked Jan 17 at 15:21
M.M.M.M.
185
asked Jan 17 at 15:21
M.M.M.M.
185
asked Jan 17 at 15:21
M.M.M.M.
185
185
1
$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36
add a comment |
1
$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36
1
1
$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36
$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36
add a comment |
1 Answer
1
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oldest
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Particular solution we find in form
$$y=e^z(Acos z+Bsin z)+Ccos z+Dsin z$$
Substitute this form in equation
$$y''-2y'+4y=e^zcos z+cos z.$$
We get
$$left( 2 B, {{e}^{z}}+3 D+2 Cright) sin z+left( 2 A, {{e}^{z}}-2 D+3 Cright) cos z={{e}^{z}} sin z+cos z,$$
$$A=0,quad B=frac12,quad C=frac{3}{13},quad D=-frac{2}{13}.$$
Then
$$y_p=frac{e^zsin z}2-frac{2sin z}{13}+frac{3cos z}{13}\=
frac{xsin(ln x)}2-frac{2sin(ln x)}{13}+frac{3cos(ln x)}{13}$$
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
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1 Answer
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active
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votes
1 Answer
1
active
oldest
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$begingroup$
Particular solution we find in form
$$y=e^z(Acos z+Bsin z)+Ccos z+Dsin z$$
Substitute this form in equation
$$y''-2y'+4y=e^zcos z+cos z.$$
We get
$$left( 2 B, {{e}^{z}}+3 D+2 Cright) sin z+left( 2 A, {{e}^{z}}-2 D+3 Cright) cos z={{e}^{z}} sin z+cos z,$$
$$A=0,quad B=frac12,quad C=frac{3}{13},quad D=-frac{2}{13}.$$
Then
$$y_p=frac{e^zsin z}2-frac{2sin z}{13}+frac{3cos z}{13}\=
frac{xsin(ln x)}2-frac{2sin(ln x)}{13}+frac{3cos(ln x)}{13}$$
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
$endgroup$
add a comment |
$begingroup$
Particular solution we find in form
$$y=e^z(Acos z+Bsin z)+Ccos z+Dsin z$$
Substitute this form in equation
$$y''-2y'+4y=e^zcos z+cos z.$$
We get
$$left( 2 B, {{e}^{z}}+3 D+2 Cright) sin z+left( 2 A, {{e}^{z}}-2 D+3 Cright) cos z={{e}^{z}} sin z+cos z,$$
$$A=0,quad B=frac12,quad C=frac{3}{13},quad D=-frac{2}{13}.$$
Then
$$y_p=frac{e^zsin z}2-frac{2sin z}{13}+frac{3cos z}{13}\=
frac{xsin(ln x)}2-frac{2sin(ln x)}{13}+frac{3cos(ln x)}{13}$$
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
$endgroup$
add a comment |
$begingroup$
Particular solution we find in form
$$y=e^z(Acos z+Bsin z)+Ccos z+Dsin z$$
Substitute this form in equation
$$y''-2y'+4y=e^zcos z+cos z.$$
We get
$$left( 2 B, {{e}^{z}}+3 D+2 Cright) sin z+left( 2 A, {{e}^{z}}-2 D+3 Cright) cos z={{e}^{z}} sin z+cos z,$$
$$A=0,quad B=frac12,quad C=frac{3}{13},quad D=-frac{2}{13}.$$
Then
$$y_p=frac{e^zsin z}2-frac{2sin z}{13}+frac{3cos z}{13}\=
frac{xsin(ln x)}2-frac{2sin(ln x)}{13}+frac{3cos(ln x)}{13}$$
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
$endgroup$
Particular solution we find in form
$$y=e^z(Acos z+Bsin z)+Ccos z+Dsin z$$
Substitute this form in equation
$$y''-2y'+4y=e^zcos z+cos z.$$
We get
$$left( 2 B, {{e}^{z}}+3 D+2 Cright) sin z+left( 2 A, {{e}^{z}}-2 D+3 Cright) cos z={{e}^{z}} sin z+cos z,$$
$$A=0,quad B=frac12,quad C=frac{3}{13},quad D=-frac{2}{13}.$$
Then
$$y_p=frac{e^zsin z}2-frac{2sin z}{13}+frac{3cos z}{13}\=
frac{xsin(ln x)}2-frac{2sin(ln x)}{13}+frac{3cos(ln x)}{13}$$
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
edited Jan 17 at 17:31
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
answered Jan 17 at 15:50
Aleksas DomarkasAleksas Domarkas
1,33716
1,33716
add a comment |
add a comment |
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$begingroup$
And now you need a particular solution for the right side $cos z+e^zsin z$. As that is not in resonance to the homogeneous solution, the method of undetermined coefficients should be easy to apply.
$endgroup$
– LutzL
Jan 17 at 15:36