Mixing
Differential equation: $y'+2xy=xe^{-2x^2}$
$begingroup$
Differential equation: $y'+2xy=xe^{-2x^2}$
Homogeneous equation is: $y=e^{-x^2}$
For non-homogeneous equation:
I started with: $y_p= dxe^{-2x^2}$ and I've got $d-2x^2d-x=0$
Then I tried with $y_p = dxe^{x^2}$ and I've got $de^{x^2}+4x^2de^{x^2}=xe^{-2x^2}$
Then I tried with $y_p=e^{x^2}$ and I've got $e^{x^2}frac{dy}{dx}=xe^{-x^2}-2xye^{x^2}$
I stopped here.
How can I solve this problem? Thank you for your help!
ordinary-differential-equations
edited Jan 11 at 12:53
Dylan
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
$endgroup$
add a comment |
$begingroup$
Differential equation: $y'+2xy=xe^{-2x^2}$
Homogeneous equation is: $y=e^{-x^2}$
For non-homogeneous equation:
I started with: $y_p= dxe^{-2x^2}$ and I've got $d-2x^2d-x=0$
Then I tried with $y_p = dxe^{x^2}$ and I've got $de^{x^2}+4x^2de^{x^2}=xe^{-2x^2}$
Then I tried with $y_p=e^{x^2}$ and I've got $e^{x^2}frac{dy}{dx}=xe^{-x^2}-2xye^{x^2}$
I stopped here.
How can I solve this problem? Thank you for your help!
ordinary-differential-equations
edited Jan 11 at 12:53
Dylan
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
$endgroup$
$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
1
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30
add a comment |
$begingroup$
Differential equation: $y'+2xy=xe^{-2x^2}$
Homogeneous equation is: $y=e^{-x^2}$
For non-homogeneous equation:
I started with: $y_p= dxe^{-2x^2}$ and I've got $d-2x^2d-x=0$
Then I tried with $y_p = dxe^{x^2}$ and I've got $de^{x^2}+4x^2de^{x^2}=xe^{-2x^2}$
Then I tried with $y_p=e^{x^2}$ and I've got $e^{x^2}frac{dy}{dx}=xe^{-x^2}-2xye^{x^2}$
I stopped here.
How can I solve this problem? Thank you for your help!
ordinary-differential-equations
edited Jan 11 at 12:53
Dylan
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
$endgroup$
Differential equation: $y'+2xy=xe^{-2x^2}$
Homogeneous equation is: $y=e^{-x^2}$
For non-homogeneous equation:
I started with: $y_p= dxe^{-2x^2}$ and I've got $d-2x^2d-x=0$
Then I tried with $y_p = dxe^{x^2}$ and I've got $de^{x^2}+4x^2de^{x^2}=xe^{-2x^2}$
Then I tried with $y_p=e^{x^2}$ and I've got $e^{x^2}frac{dy}{dx}=xe^{-x^2}-2xye^{x^2}$
I stopped here.
How can I solve this problem? Thank you for your help!
ordinary-differential-equations
ordinary-differential-equations
edited Jan 11 at 12:53
Dylan
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
edited Jan 11 at 12:53
Dylan
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
edited Jan 11 at 12:53
Dylan
12.8k31026
edited Jan 11 at 12:53
Dylan
12.8k31026
edited Jan 11 at 12:53
Dylan
12.8k31026
12.8k31026
asked Jan 10 at 9:31
J.DoeJ.Doe
899
asked Jan 10 at 9:31
J.DoeJ.Doe
899
asked Jan 10 at 9:31
J.DoeJ.Doe
899
899
$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
1
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30
add a comment |
$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
1
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30
$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
1
1
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
By product rule $frac d {dx} (e^{x^{2}}y)=e^{x^{2}}(y'+2xy)=e^{x^{2}}(xe^{-2x^{2}})=xe^{-x^{2}}$. Now integrate. The answer is $y=ce^{-x^{2}}-frac 1 2 e^{-2x^{2}}$ where $c$ is a constant. Note: $e^{x^{2}}=e^{int 2tdt}$. This function is called the integrating factor. For any equation of teh type $y'+f(x)y=g(x)$ you can multiply by the integrating factor $e^{int f(t)dt}$ to solve the equation.
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
$endgroup$
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
add a comment |
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1 Answer
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1 Answer
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$begingroup$
By product rule $frac d {dx} (e^{x^{2}}y)=e^{x^{2}}(y'+2xy)=e^{x^{2}}(xe^{-2x^{2}})=xe^{-x^{2}}$. Now integrate. The answer is $y=ce^{-x^{2}}-frac 1 2 e^{-2x^{2}}$ where $c$ is a constant. Note: $e^{x^{2}}=e^{int 2tdt}$. This function is called the integrating factor. For any equation of teh type $y'+f(x)y=g(x)$ you can multiply by the integrating factor $e^{int f(t)dt}$ to solve the equation.
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
$endgroup$
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
add a comment |
$begingroup$
By product rule $frac d {dx} (e^{x^{2}}y)=e^{x^{2}}(y'+2xy)=e^{x^{2}}(xe^{-2x^{2}})=xe^{-x^{2}}$. Now integrate. The answer is $y=ce^{-x^{2}}-frac 1 2 e^{-2x^{2}}$ where $c$ is a constant. Note: $e^{x^{2}}=e^{int 2tdt}$. This function is called the integrating factor. For any equation of teh type $y'+f(x)y=g(x)$ you can multiply by the integrating factor $e^{int f(t)dt}$ to solve the equation.
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
$endgroup$
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
add a comment |
$begingroup$
By product rule $frac d {dx} (e^{x^{2}}y)=e^{x^{2}}(y'+2xy)=e^{x^{2}}(xe^{-2x^{2}})=xe^{-x^{2}}$. Now integrate. The answer is $y=ce^{-x^{2}}-frac 1 2 e^{-2x^{2}}$ where $c$ is a constant. Note: $e^{x^{2}}=e^{int 2tdt}$. This function is called the integrating factor. For any equation of teh type $y'+f(x)y=g(x)$ you can multiply by the integrating factor $e^{int f(t)dt}$ to solve the equation.
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
$endgroup$
By product rule $frac d {dx} (e^{x^{2}}y)=e^{x^{2}}(y'+2xy)=e^{x^{2}}(xe^{-2x^{2}})=xe^{-x^{2}}$. Now integrate. The answer is $y=ce^{-x^{2}}-frac 1 2 e^{-2x^{2}}$ where $c$ is a constant. Note: $e^{x^{2}}=e^{int 2tdt}$. This function is called the integrating factor. For any equation of teh type $y'+f(x)y=g(x)$ you can multiply by the integrating factor $e^{int f(t)dt}$ to solve the equation.
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
edited Jan 10 at 9:56
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
answered Jan 10 at 9:35
Kavi Rama MurthyKavi Rama Murthy
57.8k42160
57.8k42160
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
add a comment |
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
1
1
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
$begingroup$
@LutzL Thanks for the comment.
$endgroup$
– Kavi Rama Murthy
Jan 10 at 9:53
add a comment |
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$begingroup$
What exactly is the method you are trying and why do you think it should work? $μ$ in this context is usually associated with an integrating factor.
$endgroup$
– LutzL
Jan 10 at 9:43
$begingroup$
I will replace $mu$ with $y_p$.
$endgroup$
– J.Doe
Jan 10 at 10:20
1
$begingroup$
Ok, that clarifies the "what". Now add something about "why should that work", as the method of undetermined coefficients only works for constant coefficients.
$endgroup$
– LutzL
Jan 10 at 10:30