- #1
James889
- 192
- 1
Hi, i would like some help with his one
I have
[tex]\left\{\begin{aligned}y^{\prime\prime}+10y^{\prime}+25y = 0 \\
y(1) = 0 \\
y^{\prime}(1) = 2\end{aligned} [/tex]
So the first part is simple
The general solution is on the form
[tex]-5\pm\frac{\sqrt{10^2-4\cdot25}}{2}[/tex]
it's a double root
So we have
[tex] y = Ae^{-5t}+Bte^{-5t}[/tex]
And
[tex]y^{\prime} = -5Ae^{-5t} -5Bte^{-5t}[/tex]
[tex]-5e^{-5t}(A+Bt)[/tex]
And since t = 1
[tex]A=e^5~~ B=-e^5[/tex] works for the first condition but not for the second.
Hm, how do you do this?
I have
[tex]\left\{\begin{aligned}y^{\prime\prime}+10y^{\prime}+25y = 0 \\
y(1) = 0 \\
y^{\prime}(1) = 2\end{aligned} [/tex]
So the first part is simple
The general solution is on the form
[tex]-5\pm\frac{\sqrt{10^2-4\cdot25}}{2}[/tex]
it's a double root
So we have
[tex] y = Ae^{-5t}+Bte^{-5t}[/tex]
And
[tex]y^{\prime} = -5Ae^{-5t} -5Bte^{-5t}[/tex]
[tex]-5e^{-5t}(A+Bt)[/tex]
And since t = 1
[tex]A=e^5~~ B=-e^5[/tex] works for the first condition but not for the second.
Hm, how do you do this?
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