Hello this is a solution of a differential equation i found. Please help me to understand this by EXPLAINING IT USING WORDS only. Im quite good in understanding words and not numbers. Please help. Thank you. Olfa+Gdxdtcorresponding•We can get the roots by using it's auxiliary equation1²+Gm+80- m² + am + 2m + 8 +0Jm(m+a) + 2 (m +a)=0- (m+4) (m+ 2) = 0=> either (m+2) =+ 8x = 6 sin 36"If m+2=0 ; m = -2mtado;m - 4= 5.So, the complementary solution isXect) 5.= 5.= 5 (6PM)5.=> 5ED -16D+1(601) (60+)sl(GDJ² -1²Sin 3t→ Xp(t) = 5sin 37(CD + 1) sin 3t360²-1(60+ 1) sin 3t36(-32) 15 xE5= 5 (60+1)-324-1sin 3t(6D+1)36x (19) -GD+1-325کر ان کی کار لانے والی-15 (ad Sin36ㅗladdade655n3tsin 37sin st(60 sin3t + sin 3t)+ sun 3t)-1(6. 3 cos 3t + sin3t)45Xp(t) = (18 cos 3t + sinst)65•The general solutiongiven byx (t) = xc (t) + Xp (t)-*XXL+) = (₁ ett + C²₂2e²Ĵ0۲۱= (18000 36 | singtJunnila

Here we find characteristics roots. Complementary Solution Particular solution.

Answered: dt -We can get the roots m² +Gm +8 -0…

Math

Advanced Math

dt -We can get the roots m² +Gm +8 -0 corresponcting by using it's ^ auxiliary equation

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter2: Equations And Inequalities

Section2.3: Quadratic Equations

Problem 17E

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