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Does Using an Integrating Factor give a General Solution? (Lin. ODE)
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When solving a linear ODE of the form $$y' + p(x)y = q(x)$$ does using an integrating factor, $e^int p dx$ provide us a method to get a general solution to the ODE, or is it a particular solution? Why / why not?
When using them to solve a problem, such as finding the solution to $$y' + 3(x+1)^-1y = 4(x+1)^-2$$ when using the integrating factor method, i.e exploiting the property that the integrating factor $A$ produces the identity $$(Ay)' = A(y' + py) = Aq$$ then integrating, I arrived at the solution $$y = frac2x+1 + fracC(x+1)^3$$ Where $C$ is some constant.
I also tried to solve the homogeneous version, ending up with $ y = K(x+1)^-3$, K is some constant. Furthermore, $y = frac2x+1$ is a particular solution to the inhomogeneous version. So, it seems that the integrating factor method has the particular solution + homogeneous solution = general solution built in, but why, and how?
differential-equations integrating-factor
asked Jul 28 at 4:59
Leekboi
392312
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up vote
1
down vote
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When solving a linear ODE of the form $$y' + p(x)y = q(x)$$ does using an integrating factor, $e^int p dx$ provide us a method to get a general solution to the ODE, or is it a particular solution? Why / why not?
When using them to solve a problem, such as finding the solution to $$y' + 3(x+1)^-1y = 4(x+1)^-2$$ when using the integrating factor method, i.e exploiting the property that the integrating factor $A$ produces the identity $$(Ay)' = A(y' + py) = Aq$$ then integrating, I arrived at the solution $$y = frac2x+1 + fracC(x+1)^3$$ Where $C$ is some constant.
I also tried to solve the homogeneous version, ending up with $ y = K(x+1)^-3$, K is some constant. Furthermore, $y = frac2x+1$ is a particular solution to the inhomogeneous version. So, it seems that the integrating factor method has the particular solution + homogeneous solution = general solution built in, but why, and how?
differential-equations integrating-factor
asked Jul 28 at 4:59
Leekboi
392312
$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
When solving a linear ODE of the form $$y' + p(x)y = q(x)$$ does using an integrating factor, $e^int p dx$ provide us a method to get a general solution to the ODE, or is it a particular solution? Why / why not?
When using them to solve a problem, such as finding the solution to $$y' + 3(x+1)^-1y = 4(x+1)^-2$$ when using the integrating factor method, i.e exploiting the property that the integrating factor $A$ produces the identity $$(Ay)' = A(y' + py) = Aq$$ then integrating, I arrived at the solution $$y = frac2x+1 + fracC(x+1)^3$$ Where $C$ is some constant.
I also tried to solve the homogeneous version, ending up with $ y = K(x+1)^-3$, K is some constant. Furthermore, $y = frac2x+1$ is a particular solution to the inhomogeneous version. So, it seems that the integrating factor method has the particular solution + homogeneous solution = general solution built in, but why, and how?
differential-equations integrating-factor
asked Jul 28 at 4:59
Leekboi
392312
When solving a linear ODE of the form $$y' + p(x)y = q(x)$$ does using an integrating factor, $e^int p dx$ provide us a method to get a general solution to the ODE, or is it a particular solution? Why / why not?
When using them to solve a problem, such as finding the solution to $$y' + 3(x+1)^-1y = 4(x+1)^-2$$ when using the integrating factor method, i.e exploiting the property that the integrating factor $A$ produces the identity $$(Ay)' = A(y' + py) = Aq$$ then integrating, I arrived at the solution $$y = frac2x+1 + fracC(x+1)^3$$ Where $C$ is some constant.
I also tried to solve the homogeneous version, ending up with $ y = K(x+1)^-3$, K is some constant. Furthermore, $y = frac2x+1$ is a particular solution to the inhomogeneous version. So, it seems that the integrating factor method has the particular solution + homogeneous solution = general solution built in, but why, and how?
differential-equations integrating-factor
asked Jul 28 at 4:59
Leekboi
392312
asked Jul 28 at 4:59
Leekboi
392312
asked Jul 28 at 4:59
Leekboi
392312
asked Jul 28 at 4:59
Leekboi
392312
392312
$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47
add a comment |Â
$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47
$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47
add a comment |Â
1 Answer
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oldest
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up vote
2
down vote
accepted
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=int_t_0^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.
answered Jul 28 at 5:15
Ian
65k24681
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=int_t_0^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.
answered Jul 28 at 5:15
Ian
65k24681
add a comment |Â
up vote
2
down vote
accepted
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=int_t_0^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.
answered Jul 28 at 5:15
Ian
65k24681
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=int_t_0^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.
answered Jul 28 at 5:15
Ian
65k24681
Short answer: the integrating factor method furnishes a general solution.
Long answer:
In first order linear ODE, the Green's function $G$ for a given $t_0$ satisfies the homogeneous ODE with $G(t_0)=1$. A particular solution to the inhomogeneous equation with $y_p(t_0)=0$ is $y_p(t)=int_t_0^t G(t-s) f(s) ds$. (This is non-obvious but standard, you can look up a proof.) A solution to the IVP with $y(t_0)=y_0$ is then given by $y=y_p+y_0 G$.
Now $G$ is really the reciprocal of an integrating factor (chosen with a suitable integration constant). In view of that you see that integrating after multiplying by $1/G$ results in $y/G=y_p/G+y_0$ and then you multiply by G to finally solve for $y$. This is the connection.
This Green's function idea generalizes to higher order linear ODE even though the integrating factor method does not.
answered Jul 28 at 5:15
Ian
65k24681
edited Jul 28 at 5:30
answered Jul 28 at 5:15
Ian
65k24681
answered Jul 28 at 5:15
Ian
65k24681
answered Jul 28 at 5:15
Ian
65k24681
65k24681
add a comment |Â
add a comment |Â
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$y=C_1-3ln(x+1)-frac4x+1$
– Dhamnekar Winod
Jul 28 at 5:25
Though the answer below seems sufficient I would like to point out just one case that is of first order linear differential equation in which I.E makes the equation homogeneous and allows partial differential solutions...
– Pi_die_die
Jul 28 at 5:42
@Leekboi Integrating factor method provids a general solution,not a particular solution. Particular solution can be found in Initial value problems.$:-)$
– Dhamnekar Winod
Jul 28 at 5:43
Sometimes general solutions can be manipulated to via change of arbitrary constant but that doesn't mean it isn't the general solution. In fact doesn't the general solution come built in with all the particular solutions?
– Pi_die_die
Jul 28 at 5:47