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victus00 [196]
3 years ago
11

[10 pts] b) y"+12y'+36 y = 0

Mathematics
1 answer:
yarga [219]3 years ago
4 0

Answer:

The general solution of the differential equation is:

y=c_1e^{-6t}+c_2te^{-6t}

Step-by-step explanation:

We have a second order homogeneous differential equation y''+12y'+36 y = 0

We need to find the characteristic polynomial

x^2+12x-36=0

Next, we find the roots as follows:

\mathrm{Solve\:by\:factoring}\\\\\mathrm{Rewrite\:}x^2+12x+36\mathrm{\:as\:}x^2+2x\cdot \:6+6^2\\\\\mathrm{Apply\:Perfect\:Square\:Formula}:\quad \left(a+b\right)^2=a^2+2ab+b^2\\\\\left(x+6\right)^2=0\\\\\mathrm{Solve\:}\:x+6=0:\quad x=-6

The roots of characteristic polynomial are r=-6 and s=-6

When the roots are real and equal the general solution of the differential equation is:

y=c_1e^{-6t}+c_2te^{-6t}

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