Find the general solution to the homogeneous differential equation dt2d2y+18dtdy+81y=0 The solution can be written in the form y=C1f1(t)+C2f2(t) with f1(t)= and f2(t)=

Step-by-step explanation:

Consider the differential equation . Let us propose a solution of the form where A is a non-zero constant. Then, the derivatives are calculated as follows: . Replacing those functions in the original equation, we obtain the following.

Given that A is non-zero and that the exponential function is non-zero, the following equation must be hold

Then, the only solution is r=-9. Then, one of the solutions is the function [text]y_1(t)=C_1 e^{-9t}[/tex]. Since the differential equation is a second degree equation, another solution must be given. Note that the second solution must be linear independent of the first found solution. This is achieved by multiplying the first solution by t i.e . Then, the complete solution is of the form

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)