Mixing
I need help finding the general solution to the differential equation $y''(t)+7y'(t)=-14$
up vote
2
down vote
favorite
What I've tried:
I have the inhomogeneous differential equation:
$$y''(t)+7y'(t)=-14$$
I find the particular solution to be on the form $$kt$$
by inserting the particular solution in the equation
$$(kt)''+7(kt)'=-14$$
and isolating for k, I get that:
$$k=-2$$
and therefore the particular solution is
$$y(t)-2t$$
I also need the general solution for the homogenous equation
$$y''(t)+7y'(t)=0$$
by finding the roots of the characteristic polynomial
$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$
I get the general solution:
$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$
Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by
$$y(t)=y_p(t)+y_{hom}(t)$$
Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be
$$y(t)=c_1+c_2e^{-7t}-2t$$
This is not consistent with Maple's result however
Can anyone see where I've gone wrong?
differential-equations
asked 2 days ago
Boris Grunwald
1417
add a comment |
up vote
2
down vote
favorite
What I've tried:
I have the inhomogeneous differential equation:
$$y''(t)+7y'(t)=-14$$
I find the particular solution to be on the form $$kt$$
by inserting the particular solution in the equation
$$(kt)''+7(kt)'=-14$$
and isolating for k, I get that:
$$k=-2$$
and therefore the particular solution is
$$y(t)-2t$$
I also need the general solution for the homogenous equation
$$y''(t)+7y'(t)=0$$
by finding the roots of the characteristic polynomial
$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$
I get the general solution:
$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$
Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by
$$y(t)=y_p(t)+y_{hom}(t)$$
Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be
$$y(t)=c_1+c_2e^{-7t}-2t$$
This is not consistent with Maple's result however
Can anyone see where I've gone wrong?
differential-equations
asked 2 days ago
Boris Grunwald
1417
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
What I've tried:
I have the inhomogeneous differential equation:
$$y''(t)+7y'(t)=-14$$
I find the particular solution to be on the form $$kt$$
by inserting the particular solution in the equation
$$(kt)''+7(kt)'=-14$$
and isolating for k, I get that:
$$k=-2$$
and therefore the particular solution is
$$y(t)-2t$$
I also need the general solution for the homogenous equation
$$y''(t)+7y'(t)=0$$
by finding the roots of the characteristic polynomial
$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$
I get the general solution:
$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$
Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by
$$y(t)=y_p(t)+y_{hom}(t)$$
Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be
$$y(t)=c_1+c_2e^{-7t}-2t$$
This is not consistent with Maple's result however
Can anyone see where I've gone wrong?
differential-equations
asked 2 days ago
Boris Grunwald
1417
What I've tried:
I have the inhomogeneous differential equation:
$$y''(t)+7y'(t)=-14$$
I find the particular solution to be on the form $$kt$$
by inserting the particular solution in the equation
$$(kt)''+7(kt)'=-14$$
and isolating for k, I get that:
$$k=-2$$
and therefore the particular solution is
$$y(t)-2t$$
I also need the general solution for the homogenous equation
$$y''(t)+7y'(t)=0$$
by finding the roots of the characteristic polynomial
$$z^2+7z=z(z+7)=0$$
$$z_1=0$$
$$z_2=-7$$
I get the general solution:
$$c_1e^{0t}+c_2e^{-7t}=c_1+c_2e^{-7t}$$
Now, according to my textbook, the general solution of an inhomogeneous differential equation is given by
$$y(t)=y_p(t)+y_{hom}(t)$$
Where $y_p(t)$ is the particular solution and $y_{hom}(t)$ is the general solution to the homogenous equation. Therefore I get the general solution to be
$$y(t)=c_1+c_2e^{-7t}-2t$$
This is not consistent with Maple's result however
Can anyone see where I've gone wrong?
differential-equations
differential-equations
asked 2 days ago
Boris Grunwald
1417
asked 2 days ago
Boris Grunwald
1417
asked 2 days ago
Boris Grunwald
1417
asked 2 days ago
Boris Grunwald
1417
asked 2 days ago
Boris Grunwald
1417
1417
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions
$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$
We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.
Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.
answered 2 days ago
mrtaurho
2,3691726
Thanks, that makes sense!
– Boris Grunwald
2 days ago
add a comment |
up vote
1
down vote
You went wrong when you thought what Maple wrote is different from your solution in any significant way.
Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.
answered 2 days ago
Arthur
108k7103186
add a comment |
up vote
0
down vote
As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions
$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$
We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.
Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.
answered 2 days ago
mrtaurho
2,3691726
Thanks, that makes sense!
– Boris Grunwald
2 days ago
add a comment |
up vote
1
down vote
accepted
You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions
$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$
We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.
Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.
answered 2 days ago
mrtaurho
2,3691726
Thanks, that makes sense!
– Boris Grunwald
2 days ago
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions
$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$
We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.
Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.
answered 2 days ago
mrtaurho
2,3691726
You have not gone wrong anywhere! It is just a question concerning the choice of the arbitrary constants $c_1,c_2$ and $C_1,C_2$ respectively. Therefore lets just take a brief look at the two given solutions
$$begin{align}
y_1(t)&=c_1+c_2e^{-7t}-2t\
y_2(t)&=C_2+frac{-C_1}{7}e^{-7t}-2t
end{align}$$
We can agree one the fact that the particular solution $-2t$ is the same in both $y_1(t)$ and $y_2(t)$. Therefore we do not have to worry about this term further. Now we can argue that $c_1=C_2$ since these are both arbitrary constants which have to be defined later one. Following a similiar agrumentation we can moreover say that $c_2=frac{-C_1}{7}$ hence there is no restriction regarding their value.
Therefore Maple aswell as you by yourself provided the exact same solution beside the choice of some arbitrary constants.
answered 2 days ago
mrtaurho
2,3691726
answered 2 days ago
mrtaurho
2,3691726
answered 2 days ago
mrtaurho
2,3691726
answered 2 days ago
mrtaurho
2,3691726
2,3691726
Thanks, that makes sense!
– Boris Grunwald
2 days ago
add a comment |
Thanks, that makes sense!
– Boris Grunwald
2 days ago
Thanks, that makes sense!
– Boris Grunwald
2 days ago
Thanks, that makes sense!
– Boris Grunwald
2 days ago
add a comment |
up vote
1
down vote
You went wrong when you thought what Maple wrote is different from your solution in any significant way.
Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.
answered 2 days ago
Arthur
108k7103186
add a comment |
up vote
1
down vote
You went wrong when you thought what Maple wrote is different from your solution in any significant way.
Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.
answered 2 days ago
Arthur
108k7103186
add a comment |
up vote
1
down vote
up vote
1
down vote
You went wrong when you thought what Maple wrote is different from your solution in any significant way.
Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.
answered 2 days ago
Arthur
108k7103186
You went wrong when you thought what Maple wrote is different from your solution in any significant way.
Maple has swapped the roles of $c_1$ and $c_2$ compared to you. And Maple's $c_1$ is seven times larger than your $c_2$ and has opposite sign, but since the constants are arbitrary anyways this doesn't matter. So you and Maple describe the exact same collection of functions.
answered 2 days ago
Arthur
108k7103186
answered 2 days ago
Arthur
108k7103186
answered 2 days ago
Arthur
108k7103186
answered 2 days ago
Arthur
108k7103186
108k7103186
add a comment |
add a comment |
up vote
0
down vote
As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
answered 2 days ago
Yves Daoust
121k668216
add a comment |
up vote
0
down vote
up vote
0
down vote
As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
answered 2 days ago
Yves Daoust
121k668216
As others said, your solution is right (only the expressions of the constant differ). Here is how I would solve.
First notice that $y$ is missing, and solve for $z:=y'$:
$$z'+7z=-14.$$
An obvious particular solution is the constant $z=-2$. Then with the characteristic polynomial $p+7$, you can immediately write the general solution,
$$z=ce^{-7t}-2.$$
Now to get $y$, you integrate once, and obtain
$$y=c_0e^{-7t}-2t+c_1.$$
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
answered 2 days ago
Yves Daoust
121k668216
121k668216
add a comment |
add a comment |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005544%2fi-need-help-finding-the-general-solution-to-the-differential-equation-yt7y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
