Method of undetermined coefficients clarification
For $$ddot{x}+x=epsilon$$ I found that $x(t)=c_1sin(t)+c_2cos(t)$. Next for method of undetermined coefficients we have begin{align*}
x&=At^2+Bt+C\
dot{x}&=2At+B\
ddot{x}&=2A.
end{align*} According to my solutions manual for $ddot{x}+x=epsilon$ we have $2A+At^2+Bt+C=epsilon$ or $x=epsilon$. My question is how did we just confirm that $x=epsilon$ in this case? I don't see how we made that conclusion.
differential-equations
asked Nov 26 '18 at 19:29
Peetrius
404111
add a comment |
For $$ddot{x}+x=epsilon$$ I found that $x(t)=c_1sin(t)+c_2cos(t)$. Next for method of undetermined coefficients we have begin{align*}
x&=At^2+Bt+C\
dot{x}&=2At+B\
ddot{x}&=2A.
end{align*} According to my solutions manual for $ddot{x}+x=epsilon$ we have $2A+At^2+Bt+C=epsilon$ or $x=epsilon$. My question is how did we just confirm that $x=epsilon$ in this case? I don't see how we made that conclusion.
differential-equations
asked Nov 26 '18 at 19:29
Peetrius
404111
add a comment |
For $$ddot{x}+x=epsilon$$ I found that $x(t)=c_1sin(t)+c_2cos(t)$. Next for method of undetermined coefficients we have begin{align*}
x&=At^2+Bt+C\
dot{x}&=2At+B\
ddot{x}&=2A.
end{align*} According to my solutions manual for $ddot{x}+x=epsilon$ we have $2A+At^2+Bt+C=epsilon$ or $x=epsilon$. My question is how did we just confirm that $x=epsilon$ in this case? I don't see how we made that conclusion.
differential-equations
asked Nov 26 '18 at 19:29
Peetrius
404111
For $$ddot{x}+x=epsilon$$ I found that $x(t)=c_1sin(t)+c_2cos(t)$. Next for method of undetermined coefficients we have begin{align*}
x&=At^2+Bt+C\
dot{x}&=2At+B\
ddot{x}&=2A.
end{align*} According to my solutions manual for $ddot{x}+x=epsilon$ we have $2A+At^2+Bt+C=epsilon$ or $x=epsilon$. My question is how did we just confirm that $x=epsilon$ in this case? I don't see how we made that conclusion.
differential-equations
differential-equations
asked Nov 26 '18 at 19:29
Peetrius
404111
asked Nov 26 '18 at 19:29
Peetrius
404111
asked Nov 26 '18 at 19:29
Peetrius
404111
asked Nov 26 '18 at 19:29
Peetrius
404111
asked Nov 26 '18 at 19:29
Peetrius
404111
404111
add a comment |
add a comment |
3 Answers
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You can identify polynomials by the coefficient before each degree. For example here
$A=0$ because there's no term in $t^2$, and $B=0$ because there's no term in $t$. However
$$
2A+C= epsilon
$$
But with $A=0$ we have
$C= epsilon$. Replacing
$$
x=At^2+Bt+C=0+0+epsilon
$$
answered Nov 26 '18 at 19:41
Atmos
4,742319
add a comment |
Substituting the polynomial yielded, you can see by equating coefficients, that $x=epsilon$ :
$$ddot{At^2 + Bt + C} + At^2 + Bt + C = epsilon Rightarrow begin{cases} A=0 \ B = 0 \C=epsilonend{cases}$$
Plugging in $x=epsilon$ if $epsilon$ is a constant in your equation, you simply get that
$$ddot{epsilon}+epsilon= 0 + epsilon = epsilon$$
thus your initial differential equation is true and $x=epsilon$ is indeed a solution.
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
add a comment |
Your right side is $ϵe^{0cdot t}$ and as $0$ is not a root of the characteristic polynomial $0=lambda^2+1$ of the left side, the method of undetermined coefficients tells you that the polynomial factor of the particular solution has the same degree $0$ with no extra factors $t$, so $x_p(t)=Ae^{0cdot t}=A$. Inserting into the ODE gives $A=ϵ$, thus $x_p(t)=ϵ$.
answered Nov 26 '18 at 20:52
LutzL
56k42054
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You can identify polynomials by the coefficient before each degree. For example here
$A=0$ because there's no term in $t^2$, and $B=0$ because there's no term in $t$. However
$$
2A+C= epsilon
$$
But with $A=0$ we have
$C= epsilon$. Replacing
$$
x=At^2+Bt+C=0+0+epsilon
$$
answered Nov 26 '18 at 19:41
Atmos
4,742319
add a comment |
You can identify polynomials by the coefficient before each degree. For example here
$A=0$ because there's no term in $t^2$, and $B=0$ because there's no term in $t$. However
$$
2A+C= epsilon
$$
But with $A=0$ we have
$C= epsilon$. Replacing
$$
x=At^2+Bt+C=0+0+epsilon
$$
answered Nov 26 '18 at 19:41
Atmos
4,742319
add a comment |
You can identify polynomials by the coefficient before each degree. For example here
$A=0$ because there's no term in $t^2$, and $B=0$ because there's no term in $t$. However
$$
2A+C= epsilon
$$
But with $A=0$ we have
$C= epsilon$. Replacing
$$
x=At^2+Bt+C=0+0+epsilon
$$
answered Nov 26 '18 at 19:41
Atmos
4,742319
You can identify polynomials by the coefficient before each degree. For example here
$A=0$ because there's no term in $t^2$, and $B=0$ because there's no term in $t$. However
$$
2A+C= epsilon
$$
But with $A=0$ we have
$C= epsilon$. Replacing
$$
x=At^2+Bt+C=0+0+epsilon
$$
answered Nov 26 '18 at 19:41
Atmos
4,742319
answered Nov 26 '18 at 19:41
Atmos
4,742319
answered Nov 26 '18 at 19:41
Atmos
4,742319
answered Nov 26 '18 at 19:41
Atmos
4,742319
4,742319
add a comment |
add a comment |
Substituting the polynomial yielded, you can see by equating coefficients, that $x=epsilon$ :
$$ddot{At^2 + Bt + C} + At^2 + Bt + C = epsilon Rightarrow begin{cases} A=0 \ B = 0 \C=epsilonend{cases}$$
Plugging in $x=epsilon$ if $epsilon$ is a constant in your equation, you simply get that
$$ddot{epsilon}+epsilon= 0 + epsilon = epsilon$$
thus your initial differential equation is true and $x=epsilon$ is indeed a solution.
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
add a comment |
Substituting the polynomial yielded, you can see by equating coefficients, that $x=epsilon$ :
$$ddot{At^2 + Bt + C} + At^2 + Bt + C = epsilon Rightarrow begin{cases} A=0 \ B = 0 \C=epsilonend{cases}$$
Plugging in $x=epsilon$ if $epsilon$ is a constant in your equation, you simply get that
$$ddot{epsilon}+epsilon= 0 + epsilon = epsilon$$
thus your initial differential equation is true and $x=epsilon$ is indeed a solution.
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
add a comment |
Substituting the polynomial yielded, you can see by equating coefficients, that $x=epsilon$ :
$$ddot{At^2 + Bt + C} + At^2 + Bt + C = epsilon Rightarrow begin{cases} A=0 \ B = 0 \C=epsilonend{cases}$$
Plugging in $x=epsilon$ if $epsilon$ is a constant in your equation, you simply get that
$$ddot{epsilon}+epsilon= 0 + epsilon = epsilon$$
thus your initial differential equation is true and $x=epsilon$ is indeed a solution.
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
Substituting the polynomial yielded, you can see by equating coefficients, that $x=epsilon$ :
$$ddot{At^2 + Bt + C} + At^2 + Bt + C = epsilon Rightarrow begin{cases} A=0 \ B = 0 \C=epsilonend{cases}$$
Plugging in $x=epsilon$ if $epsilon$ is a constant in your equation, you simply get that
$$ddot{epsilon}+epsilon= 0 + epsilon = epsilon$$
thus your initial differential equation is true and $x=epsilon$ is indeed a solution.
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
answered Nov 26 '18 at 19:40
Rebellos
14.4k31245
14.4k31245
add a comment |
add a comment |
Your right side is $ϵe^{0cdot t}$ and as $0$ is not a root of the characteristic polynomial $0=lambda^2+1$ of the left side, the method of undetermined coefficients tells you that the polynomial factor of the particular solution has the same degree $0$ with no extra factors $t$, so $x_p(t)=Ae^{0cdot t}=A$. Inserting into the ODE gives $A=ϵ$, thus $x_p(t)=ϵ$.
answered Nov 26 '18 at 20:52
LutzL
56k42054
add a comment |
Your right side is $ϵe^{0cdot t}$ and as $0$ is not a root of the characteristic polynomial $0=lambda^2+1$ of the left side, the method of undetermined coefficients tells you that the polynomial factor of the particular solution has the same degree $0$ with no extra factors $t$, so $x_p(t)=Ae^{0cdot t}=A$. Inserting into the ODE gives $A=ϵ$, thus $x_p(t)=ϵ$.
answered Nov 26 '18 at 20:52
LutzL
56k42054
add a comment |
Your right side is $ϵe^{0cdot t}$ and as $0$ is not a root of the characteristic polynomial $0=lambda^2+1$ of the left side, the method of undetermined coefficients tells you that the polynomial factor of the particular solution has the same degree $0$ with no extra factors $t$, so $x_p(t)=Ae^{0cdot t}=A$. Inserting into the ODE gives $A=ϵ$, thus $x_p(t)=ϵ$.
answered Nov 26 '18 at 20:52
LutzL
56k42054
Your right side is $ϵe^{0cdot t}$ and as $0$ is not a root of the characteristic polynomial $0=lambda^2+1$ of the left side, the method of undetermined coefficients tells you that the polynomial factor of the particular solution has the same degree $0$ with no extra factors $t$, so $x_p(t)=Ae^{0cdot t}=A$. Inserting into the ODE gives $A=ϵ$, thus $x_p(t)=ϵ$.
answered Nov 26 '18 at 20:52
LutzL
56k42054
answered Nov 26 '18 at 20:52
LutzL
56k42054
answered Nov 26 '18 at 20:52
LutzL
56k42054
answered Nov 26 '18 at 20:52
LutzL
56k42054
56k42054
add a comment |
add a comment |
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