Solving ordinary differential equations of order 2
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I'm struggling solving this linear ODE:
$$y''(x) + 2y'(x) = -4$$
I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;
But I'm struggling with the non-homogeneous part:
If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.
What am I doing wrong! Thanks a lot!
real-analysis ordinary-differential-equations
edited Jan 20 at 9:15
Robert Z
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
$endgroup$
add a comment |
$begingroup$
I'm struggling solving this linear ODE:
$$y''(x) + 2y'(x) = -4$$
I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;
But I'm struggling with the non-homogeneous part:
If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.
What am I doing wrong! Thanks a lot!
real-analysis ordinary-differential-equations
edited Jan 20 at 9:15
Robert Z
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
$endgroup$
$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38
add a comment |
$begingroup$
I'm struggling solving this linear ODE:
$$y''(x) + 2y'(x) = -4$$
I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;
But I'm struggling with the non-homogeneous part:
If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.
What am I doing wrong! Thanks a lot!
real-analysis ordinary-differential-equations
edited Jan 20 at 9:15
Robert Z
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
$endgroup$
I'm struggling solving this linear ODE:
$$y''(x) + 2y'(x) = -4$$
I solved the homogeneous solution for this equation: $c_1 + c_2 e^{-2x}$ for some constants $c_1$, $c_2$;
But I'm struggling with the non-homogeneous part:
If I choose $y_p(x) = C$ for some constant $C$, then $y_p' = 0$, $y_p'' = 0$. Plugging this equations into the original one, this would yield: $0 = -4$.
What am I doing wrong! Thanks a lot!
real-analysis ordinary-differential-equations
real-analysis ordinary-differential-equations
edited Jan 20 at 9:15
Robert Z
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
edited Jan 20 at 9:15
Robert Z
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
edited Jan 20 at 9:15
Robert Z
98.3k1067139
edited Jan 20 at 9:15
Robert Z
98.3k1067139
edited Jan 20 at 9:15
Robert Z
98.3k1067139
98.3k1067139
asked Jan 20 at 9:07
scalpulascalpula
124
asked Jan 20 at 9:07
scalpulascalpula
124
asked Jan 20 at 9:07
scalpulascalpula
124
124
$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38
add a comment |
$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38
$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38
$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=xcdot C$$
where $C$ is a constant to be determined.
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
$endgroup$
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
$begingroup$
Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=xcdot C$$
where $C$ is a constant to be determined.
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
$endgroup$
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
add a comment |
$begingroup$
Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=xcdot C$$
where $C$ is a constant to be determined.
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
$endgroup$
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
add a comment |
$begingroup$
Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=xcdot C$$
where $C$ is a constant to be determined.
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
$endgroup$
Hint. Since $0$ is a solution of multiplicity $1$ of the characteristic equation $z^2+2z=0$ and $-4$ is a polynomial of zero degree then, by the Method of undetermined coefficients, you should try as a particular solution the following form
$$y_p(x)=xcdot C$$
where $C$ is a constant to be determined.
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
edited Jan 20 at 9:16
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
answered Jan 20 at 9:10
Robert ZRobert Z
98.3k1067139
98.3k1067139
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
add a comment |
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
$begingroup$
Thanks a lot! I totolly forgot that I have to multiply C with x . Now it makes sense! :)
$endgroup$
– scalpula
Jan 20 at 12:11
add a comment |
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$begingroup$
You can follow the hint of Robert Z. For this particular second-order ODE it might be easier to reduce it to first order, as $y$ does not appear: define $z := y'$ to obtain a first-order linear ODE for $z$. Solve for $z$, then integrate once to obtain $y$.
$endgroup$
– Christoph
Jan 20 at 9:38