Help solving a system of differential equations
$begingroup$
Consider the third order linear differential equation:
$x'''- 2x''- 4x'+ 8x = 0.$
Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.
I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:
$x_1' = x_2$,
$x_2' = x_3$,
$x_3' = -8x_1+4x_2+2x_3$.
I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.
linear-algebra ordinary-differential-equations
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
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add a comment |
$begingroup$
Consider the third order linear differential equation:
$x'''- 2x''- 4x'+ 8x = 0.$
Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.
I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:
$x_1' = x_2$,
$x_2' = x_3$,
$x_3' = -8x_1+4x_2+2x_3$.
I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.
linear-algebra ordinary-differential-equations
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
$endgroup$
$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48
add a comment |
$begingroup$
Consider the third order linear differential equation:
$x'''- 2x''- 4x'+ 8x = 0.$
Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.
I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:
$x_1' = x_2$,
$x_2' = x_3$,
$x_3' = -8x_1+4x_2+2x_3$.
I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.
linear-algebra ordinary-differential-equations
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
$endgroup$
Consider the third order linear differential equation:
$x'''- 2x''- 4x'+ 8x = 0.$
Given that $x(0) = 4$, $x'(0) = 16$, and $x''(0) = 16$, find the general solution to the corresponding first-order system.
I know that if I set $x_1 = x$, $x_2 = x_1'$, and $x_3 = x_2'$, I can generate the following system of equations:
$x_1' = x_2$,
$x_2' = x_3$,
$x_3' = -8x_1+4x_2+2x_3$.
I have got 2 (multiplicity of 2) and -2 (multiplicity of 1) for eigenvalues, but I'm not sure how to finish from here. Help would be appreciated.
linear-algebra ordinary-differential-equations
linear-algebra ordinary-differential-equations
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
edited Nov 29 '18 at 9:47
realgneel
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
asked Nov 29 '18 at 8:31
realgneelrealgneel
32
32
$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48
add a comment |
$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48
$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48
$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48
add a comment |
2 Answers
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oldest
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$begingroup$
I think you have an error in your equation. Looking at your expression for $x_3$ it should be:
$$x'''- 2x''- 4x'+ 8x = 0$$
...and the characteristic equation is:
$$r^3-2r^2-4r+8=0$$
$$(r^3+8)-(2r^2+4r)=0$$
$$(r+2)(r^2-2r+4) -2r(r+2)=0$$
$$(r+2)(r^2-4r+4)=0$$
$$(r+2)(r-2)^2=0$$
$$r_1=-2, quad r_{2,3}=2$$
So the solution to your equation is:
$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$
Constants can be evaluated from your initial conditions. You should be able to proceed from here.
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
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$begingroup$
I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.
The general solution is
$x(t)=Ae^{-2t} +(Bt+C)e^{2t}$
for constants $A, B, C$. So we have
$x(0) = A+C = 4$
$x'(0) = -2A + B +2C = 16$
$x''(0) = 4A + 4B + 4C = 16$
Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
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2 Answers
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2 Answers
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$begingroup$
I think you have an error in your equation. Looking at your expression for $x_3$ it should be:
$$x'''- 2x''- 4x'+ 8x = 0$$
...and the characteristic equation is:
$$r^3-2r^2-4r+8=0$$
$$(r^3+8)-(2r^2+4r)=0$$
$$(r+2)(r^2-2r+4) -2r(r+2)=0$$
$$(r+2)(r^2-4r+4)=0$$
$$(r+2)(r-2)^2=0$$
$$r_1=-2, quad r_{2,3}=2$$
So the solution to your equation is:
$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$
Constants can be evaluated from your initial conditions. You should be able to proceed from here.
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
$endgroup$
add a comment |
$begingroup$
I think you have an error in your equation. Looking at your expression for $x_3$ it should be:
$$x'''- 2x''- 4x'+ 8x = 0$$
...and the characteristic equation is:
$$r^3-2r^2-4r+8=0$$
$$(r^3+8)-(2r^2+4r)=0$$
$$(r+2)(r^2-2r+4) -2r(r+2)=0$$
$$(r+2)(r^2-4r+4)=0$$
$$(r+2)(r-2)^2=0$$
$$r_1=-2, quad r_{2,3}=2$$
So the solution to your equation is:
$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$
Constants can be evaluated from your initial conditions. You should be able to proceed from here.
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
$endgroup$
add a comment |
$begingroup$
I think you have an error in your equation. Looking at your expression for $x_3$ it should be:
$$x'''- 2x''- 4x'+ 8x = 0$$
...and the characteristic equation is:
$$r^3-2r^2-4r+8=0$$
$$(r^3+8)-(2r^2+4r)=0$$
$$(r+2)(r^2-2r+4) -2r(r+2)=0$$
$$(r+2)(r^2-4r+4)=0$$
$$(r+2)(r-2)^2=0$$
$$r_1=-2, quad r_{2,3}=2$$
So the solution to your equation is:
$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$
Constants can be evaluated from your initial conditions. You should be able to proceed from here.
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
$endgroup$
I think you have an error in your equation. Looking at your expression for $x_3$ it should be:
$$x'''- 2x''- 4x'+ 8x = 0$$
...and the characteristic equation is:
$$r^3-2r^2-4r+8=0$$
$$(r^3+8)-(2r^2+4r)=0$$
$$(r+2)(r^2-2r+4) -2r(r+2)=0$$
$$(r+2)(r^2-4r+4)=0$$
$$(r+2)(r-2)^2=0$$
$$r_1=-2, quad r_{2,3}=2$$
So the solution to your equation is:
$$x(t)=C_1e^{-2t}+C_2e^{2t}+C_3te^{2t}$$
Constants can be evaluated from your initial conditions. You should be able to proceed from here.
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
answered Nov 29 '18 at 9:00
OldboyOldboy
7,9301935
7,9301935
add a comment |
add a comment |
$begingroup$
I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.
The general solution is
$x(t)=Ae^{-2t} +(Bt+C)e^{2t}$
for constants $A, B, C$. So we have
$x(0) = A+C = 4$
$x'(0) = -2A + B +2C = 16$
$x''(0) = 4A + 4B + 4C = 16$
Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
$endgroup$
add a comment |
$begingroup$
I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.
The general solution is
$x(t)=Ae^{-2t} +(Bt+C)e^{2t}$
for constants $A, B, C$. So we have
$x(0) = A+C = 4$
$x'(0) = -2A + B +2C = 16$
$x''(0) = 4A + 4B + 4C = 16$
Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
$endgroup$
add a comment |
$begingroup$
I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.
The general solution is
$x(t)=Ae^{-2t} +(Bt+C)e^{2t}$
for constants $A, B, C$. So we have
$x(0) = A+C = 4$
$x'(0) = -2A + B +2C = 16$
$x''(0) = 4A + 4B + 4C = 16$
Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
$endgroup$
I will assume the equation is meant to be $x'''-2x''-4x'+8x=0$, in which case its characteristic polynomial is indeed $(r-2)^2(r+2)$ and the rest of the post makes sense.
The general solution is
$x(t)=Ae^{-2t} +(Bt+C)e^{2t}$
for constants $A, B, C$. So we have
$x(0) = A+C = 4$
$x'(0) = -2A + B +2C = 16$
$x''(0) = 4A + 4B + 4C = 16$
Now solve these simultaneous equations in $A,B,C$ to find the specific solution that meets your boundary conditions.
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
answered Nov 29 '18 at 9:08
gandalf61gandalf61
8,536725
8,536725
add a comment |
add a comment |
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$begingroup$
Yes, thanks for pointing that out, I wrote a typo.
$endgroup$
– realgneel
Nov 29 '18 at 9:48