Method of Undermined coefficients
$begingroup$
The following equation:
$y''+y=cos(x)$
Solving $y''+y=0$ gives me $c_1cos(x)+c_2sin(x)$
Using the particular integral yp form of $acos(x)+bsin(x)$ and substituting this back into the ODE
$y = acos(x)+bsin(x)$, $y'=-asin(x)+bcos(x)$, $y''=-acos(x)-bsin(x)$
Plugging this back into the ODE leaves me with $0=cos(x)$
So how is the particular solution $y=1/2xsin(x)$ with the coefficients value of $c_1=0$ and $c_2=1/2$, and where does the $x$ come from?
calculus multivariable-calculus numerical-methods
edited Dec 19 '18 at 18:09
Andrei
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
$endgroup$
add a comment |
$begingroup$
The following equation:
$y''+y=cos(x)$
Solving $y''+y=0$ gives me $c_1cos(x)+c_2sin(x)$
Using the particular integral yp form of $acos(x)+bsin(x)$ and substituting this back into the ODE
$y = acos(x)+bsin(x)$, $y'=-asin(x)+bcos(x)$, $y''=-acos(x)-bsin(x)$
Plugging this back into the ODE leaves me with $0=cos(x)$
So how is the particular solution $y=1/2xsin(x)$ with the coefficients value of $c_1=0$ and $c_2=1/2$, and where does the $x$ come from?
calculus multivariable-calculus numerical-methods
edited Dec 19 '18 at 18:09
Andrei
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
$endgroup$
2
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
2
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02
add a comment |
$begingroup$
The following equation:
$y''+y=cos(x)$
Solving $y''+y=0$ gives me $c_1cos(x)+c_2sin(x)$
Using the particular integral yp form of $acos(x)+bsin(x)$ and substituting this back into the ODE
$y = acos(x)+bsin(x)$, $y'=-asin(x)+bcos(x)$, $y''=-acos(x)-bsin(x)$
Plugging this back into the ODE leaves me with $0=cos(x)$
So how is the particular solution $y=1/2xsin(x)$ with the coefficients value of $c_1=0$ and $c_2=1/2$, and where does the $x$ come from?
calculus multivariable-calculus numerical-methods
edited Dec 19 '18 at 18:09
Andrei
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
$endgroup$
The following equation:
$y''+y=cos(x)$
Solving $y''+y=0$ gives me $c_1cos(x)+c_2sin(x)$
Using the particular integral yp form of $acos(x)+bsin(x)$ and substituting this back into the ODE
$y = acos(x)+bsin(x)$, $y'=-asin(x)+bcos(x)$, $y''=-acos(x)-bsin(x)$
Plugging this back into the ODE leaves me with $0=cos(x)$
So how is the particular solution $y=1/2xsin(x)$ with the coefficients value of $c_1=0$ and $c_2=1/2$, and where does the $x$ come from?
calculus multivariable-calculus numerical-methods
calculus multivariable-calculus numerical-methods
edited Dec 19 '18 at 18:09
Andrei
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
edited Dec 19 '18 at 18:09
Andrei
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
edited Dec 19 '18 at 18:09
Andrei
12k21126
edited Dec 19 '18 at 18:09
Andrei
12k21126
edited Dec 19 '18 at 18:09
Andrei
12k21126
12k21126
asked Dec 19 '18 at 17:52
luffyluffy
92
asked Dec 19 '18 at 17:52
luffyluffy
92
asked Dec 19 '18 at 17:52
luffyluffy
92
92
2
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
2
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02
add a comment |
2
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
2
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02
2
2
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
2
2
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Observe that $y''+y = 0 Rightarrow y=c_1 cos(x) + c_2 sin(x)$. So your forcing term $f(x) = cos(x)$ is a solution to the homogenous solution, so we need to come up with a new basis for the method of undetermined coefficients. Since equating coefficients will fail for this basis since if $y=c_1 cos(x) + c_2 sin(x)$, the left hand side will always be a $0$ sicne it's a solution to the homogenous equation .The main way is to differentiate the homogenous solution $y_h$ with respect to $r$ where $r$ is the root of the characteristic polynomial. For this case,
$y(x) = c_1e^{rx} + c_2e^{-rx}, r = i$, so $frac{dy}{dr} = xy_h(x) = x(alpha_1cos(x) + alpha_2sin(x))$ after the usual linear combination rearrangement trick for $e^{ix},e^{-ix}$ using Euler's identity. So our new basis for trial solutions $x(alpha_1cos(x) + alpha_2sin(x))$ which after equating coefficients will give you $alpha_1 = 0, alpha_2 = frac{1}{2}$. Remember to add in the homogenous solution at the end.
answered Dec 19 '18 at 18:16
Story123Story123
22017
$endgroup$
add a comment |
$begingroup$
Let's choose a more general particular solution for $y''+y=cos(nx)$ in the form $y_p=(ax+b)sin(nx)+(cx+d)cos(nx)$, with $n>0$. You can easily show
$$y_p''=−n((anx+bn+2c)sin(nx)+(cnx+dn−2a)cos(nx))$$
Let's plug these two into the original equation, and you get $$xsin(nx)(-an^2+a)+sin(nx)(-bn^2-2nc+b)+xcos(nx)(-cn^2+c)+cos(nx)(-dn^2+2an+d)=cos(nx)$$
In case $n=1$, you have $1-n^2=0$, so your equation simplifies to $$-2csin(x)+2acos(x)=cos(x)$$
The solutions are therefore $c=0$ and $a=1/2$, so your solution is $$y_p=frac 12xsin s$$
It is interesting to see what happens if $nne 1$. Then if you kook at just the coefficients of $xsin(nx)$ you get $a(1-n^2)=0$. That means $a=0$. Similarly, for $xcos(nx)$ you get $c=0$. You equation reduces to $$b(1-n^2)sin(nx)+d(1-n^2)cos(nx)=cos(nx)$$
You have therefore $b=0$ and $d=frac1{1-n^2}$. So no terms with $x$ multiplied with a trigonometric function.
So the conclusion is that if the free term in the ODE is the same type as the homogeneous solution, the particular solution is the form $axsin x+bxcos x$, otherwise it does not contain $x$.
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
$endgroup$
add a comment |
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Observe that $y''+y = 0 Rightarrow y=c_1 cos(x) + c_2 sin(x)$. So your forcing term $f(x) = cos(x)$ is a solution to the homogenous solution, so we need to come up with a new basis for the method of undetermined coefficients. Since equating coefficients will fail for this basis since if $y=c_1 cos(x) + c_2 sin(x)$, the left hand side will always be a $0$ sicne it's a solution to the homogenous equation .The main way is to differentiate the homogenous solution $y_h$ with respect to $r$ where $r$ is the root of the characteristic polynomial. For this case,
$y(x) = c_1e^{rx} + c_2e^{-rx}, r = i$, so $frac{dy}{dr} = xy_h(x) = x(alpha_1cos(x) + alpha_2sin(x))$ after the usual linear combination rearrangement trick for $e^{ix},e^{-ix}$ using Euler's identity. So our new basis for trial solutions $x(alpha_1cos(x) + alpha_2sin(x))$ which after equating coefficients will give you $alpha_1 = 0, alpha_2 = frac{1}{2}$. Remember to add in the homogenous solution at the end.
answered Dec 19 '18 at 18:16
Story123Story123
22017
$endgroup$
add a comment |
$begingroup$
Observe that $y''+y = 0 Rightarrow y=c_1 cos(x) + c_2 sin(x)$. So your forcing term $f(x) = cos(x)$ is a solution to the homogenous solution, so we need to come up with a new basis for the method of undetermined coefficients. Since equating coefficients will fail for this basis since if $y=c_1 cos(x) + c_2 sin(x)$, the left hand side will always be a $0$ sicne it's a solution to the homogenous equation .The main way is to differentiate the homogenous solution $y_h$ with respect to $r$ where $r$ is the root of the characteristic polynomial. For this case,
$y(x) = c_1e^{rx} + c_2e^{-rx}, r = i$, so $frac{dy}{dr} = xy_h(x) = x(alpha_1cos(x) + alpha_2sin(x))$ after the usual linear combination rearrangement trick for $e^{ix},e^{-ix}$ using Euler's identity. So our new basis for trial solutions $x(alpha_1cos(x) + alpha_2sin(x))$ which after equating coefficients will give you $alpha_1 = 0, alpha_2 = frac{1}{2}$. Remember to add in the homogenous solution at the end.
answered Dec 19 '18 at 18:16
Story123Story123
22017
$endgroup$
add a comment |
$begingroup$
Observe that $y''+y = 0 Rightarrow y=c_1 cos(x) + c_2 sin(x)$. So your forcing term $f(x) = cos(x)$ is a solution to the homogenous solution, so we need to come up with a new basis for the method of undetermined coefficients. Since equating coefficients will fail for this basis since if $y=c_1 cos(x) + c_2 sin(x)$, the left hand side will always be a $0$ sicne it's a solution to the homogenous equation .The main way is to differentiate the homogenous solution $y_h$ with respect to $r$ where $r$ is the root of the characteristic polynomial. For this case,
$y(x) = c_1e^{rx} + c_2e^{-rx}, r = i$, so $frac{dy}{dr} = xy_h(x) = x(alpha_1cos(x) + alpha_2sin(x))$ after the usual linear combination rearrangement trick for $e^{ix},e^{-ix}$ using Euler's identity. So our new basis for trial solutions $x(alpha_1cos(x) + alpha_2sin(x))$ which after equating coefficients will give you $alpha_1 = 0, alpha_2 = frac{1}{2}$. Remember to add in the homogenous solution at the end.
answered Dec 19 '18 at 18:16
Story123Story123
22017
$endgroup$
Observe that $y''+y = 0 Rightarrow y=c_1 cos(x) + c_2 sin(x)$. So your forcing term $f(x) = cos(x)$ is a solution to the homogenous solution, so we need to come up with a new basis for the method of undetermined coefficients. Since equating coefficients will fail for this basis since if $y=c_1 cos(x) + c_2 sin(x)$, the left hand side will always be a $0$ sicne it's a solution to the homogenous equation .The main way is to differentiate the homogenous solution $y_h$ with respect to $r$ where $r$ is the root of the characteristic polynomial. For this case,
$y(x) = c_1e^{rx} + c_2e^{-rx}, r = i$, so $frac{dy}{dr} = xy_h(x) = x(alpha_1cos(x) + alpha_2sin(x))$ after the usual linear combination rearrangement trick for $e^{ix},e^{-ix}$ using Euler's identity. So our new basis for trial solutions $x(alpha_1cos(x) + alpha_2sin(x))$ which after equating coefficients will give you $alpha_1 = 0, alpha_2 = frac{1}{2}$. Remember to add in the homogenous solution at the end.
answered Dec 19 '18 at 18:16
Story123Story123
22017
answered Dec 19 '18 at 18:16
Story123Story123
22017
answered Dec 19 '18 at 18:16
Story123Story123
22017
answered Dec 19 '18 at 18:16
Story123Story123
22017
22017
add a comment |
add a comment |
$begingroup$
Let's choose a more general particular solution for $y''+y=cos(nx)$ in the form $y_p=(ax+b)sin(nx)+(cx+d)cos(nx)$, with $n>0$. You can easily show
$$y_p''=−n((anx+bn+2c)sin(nx)+(cnx+dn−2a)cos(nx))$$
Let's plug these two into the original equation, and you get $$xsin(nx)(-an^2+a)+sin(nx)(-bn^2-2nc+b)+xcos(nx)(-cn^2+c)+cos(nx)(-dn^2+2an+d)=cos(nx)$$
In case $n=1$, you have $1-n^2=0$, so your equation simplifies to $$-2csin(x)+2acos(x)=cos(x)$$
The solutions are therefore $c=0$ and $a=1/2$, so your solution is $$y_p=frac 12xsin s$$
It is interesting to see what happens if $nne 1$. Then if you kook at just the coefficients of $xsin(nx)$ you get $a(1-n^2)=0$. That means $a=0$. Similarly, for $xcos(nx)$ you get $c=0$. You equation reduces to $$b(1-n^2)sin(nx)+d(1-n^2)cos(nx)=cos(nx)$$
You have therefore $b=0$ and $d=frac1{1-n^2}$. So no terms with $x$ multiplied with a trigonometric function.
So the conclusion is that if the free term in the ODE is the same type as the homogeneous solution, the particular solution is the form $axsin x+bxcos x$, otherwise it does not contain $x$.
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
$endgroup$
add a comment |
$begingroup$
Let's choose a more general particular solution for $y''+y=cos(nx)$ in the form $y_p=(ax+b)sin(nx)+(cx+d)cos(nx)$, with $n>0$. You can easily show
$$y_p''=−n((anx+bn+2c)sin(nx)+(cnx+dn−2a)cos(nx))$$
Let's plug these two into the original equation, and you get $$xsin(nx)(-an^2+a)+sin(nx)(-bn^2-2nc+b)+xcos(nx)(-cn^2+c)+cos(nx)(-dn^2+2an+d)=cos(nx)$$
In case $n=1$, you have $1-n^2=0$, so your equation simplifies to $$-2csin(x)+2acos(x)=cos(x)$$
The solutions are therefore $c=0$ and $a=1/2$, so your solution is $$y_p=frac 12xsin s$$
It is interesting to see what happens if $nne 1$. Then if you kook at just the coefficients of $xsin(nx)$ you get $a(1-n^2)=0$. That means $a=0$. Similarly, for $xcos(nx)$ you get $c=0$. You equation reduces to $$b(1-n^2)sin(nx)+d(1-n^2)cos(nx)=cos(nx)$$
You have therefore $b=0$ and $d=frac1{1-n^2}$. So no terms with $x$ multiplied with a trigonometric function.
So the conclusion is that if the free term in the ODE is the same type as the homogeneous solution, the particular solution is the form $axsin x+bxcos x$, otherwise it does not contain $x$.
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
$endgroup$
add a comment |
$begingroup$
Let's choose a more general particular solution for $y''+y=cos(nx)$ in the form $y_p=(ax+b)sin(nx)+(cx+d)cos(nx)$, with $n>0$. You can easily show
$$y_p''=−n((anx+bn+2c)sin(nx)+(cnx+dn−2a)cos(nx))$$
Let's plug these two into the original equation, and you get $$xsin(nx)(-an^2+a)+sin(nx)(-bn^2-2nc+b)+xcos(nx)(-cn^2+c)+cos(nx)(-dn^2+2an+d)=cos(nx)$$
In case $n=1$, you have $1-n^2=0$, so your equation simplifies to $$-2csin(x)+2acos(x)=cos(x)$$
The solutions are therefore $c=0$ and $a=1/2$, so your solution is $$y_p=frac 12xsin s$$
It is interesting to see what happens if $nne 1$. Then if you kook at just the coefficients of $xsin(nx)$ you get $a(1-n^2)=0$. That means $a=0$. Similarly, for $xcos(nx)$ you get $c=0$. You equation reduces to $$b(1-n^2)sin(nx)+d(1-n^2)cos(nx)=cos(nx)$$
You have therefore $b=0$ and $d=frac1{1-n^2}$. So no terms with $x$ multiplied with a trigonometric function.
So the conclusion is that if the free term in the ODE is the same type as the homogeneous solution, the particular solution is the form $axsin x+bxcos x$, otherwise it does not contain $x$.
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
$endgroup$
Let's choose a more general particular solution for $y''+y=cos(nx)$ in the form $y_p=(ax+b)sin(nx)+(cx+d)cos(nx)$, with $n>0$. You can easily show
$$y_p''=−n((anx+bn+2c)sin(nx)+(cnx+dn−2a)cos(nx))$$
Let's plug these two into the original equation, and you get $$xsin(nx)(-an^2+a)+sin(nx)(-bn^2-2nc+b)+xcos(nx)(-cn^2+c)+cos(nx)(-dn^2+2an+d)=cos(nx)$$
In case $n=1$, you have $1-n^2=0$, so your equation simplifies to $$-2csin(x)+2acos(x)=cos(x)$$
The solutions are therefore $c=0$ and $a=1/2$, so your solution is $$y_p=frac 12xsin s$$
It is interesting to see what happens if $nne 1$. Then if you kook at just the coefficients of $xsin(nx)$ you get $a(1-n^2)=0$. That means $a=0$. Similarly, for $xcos(nx)$ you get $c=0$. You equation reduces to $$b(1-n^2)sin(nx)+d(1-n^2)cos(nx)=cos(nx)$$
You have therefore $b=0$ and $d=frac1{1-n^2}$. So no terms with $x$ multiplied with a trigonometric function.
So the conclusion is that if the free term in the ODE is the same type as the homogeneous solution, the particular solution is the form $axsin x+bxcos x$, otherwise it does not contain $x$.
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
answered Dec 19 '18 at 18:50
AndreiAndrei
12k21126
12k21126
add a comment |
add a comment |
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2
$begingroup$
The problem you have is that the particular solution is a form of the homogeneous solution. You need to multiply your particular solution by $x$ so as to avoid a homogenous solution.
$endgroup$
– Joel Pereira
Dec 19 '18 at 17:58
$begingroup$
Okay, but how is the coefficient c1=0 and c2=1/2 when the $cos(x), sin(x)$ all cancel out which leaves $0=cos(x)$
$endgroup$
– luffy
Dec 19 '18 at 18:01
2
$begingroup$
You let $y_p = axcos(x)+bxsin(x)$. Now find $y^{,prime}_p$ and $y_p^{,primeprime}$ and plug into your equation. You will be able to find the coefficients. Be careful: the derivatives involve the product rule.
$endgroup$
– Joel Pereira
Dec 19 '18 at 18:02