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Find a general solution for t<0.t2y''(t)+9ty'(t)+17y(t)=0

Short Answer

Expert verified

The general solution of the given equation t2y''(t)+9ty'(t)+17y(t)=0is y(t)=c1t-4cos(ln(-t))+c2t-4sin(ln(-t))for t<0.

Step by step solution

01

Find the associated characteristic equation.

The associated characteristic equation to at2y''(t)+bty'(t)+cy(t)=0 be equationr2+(b-a)r+c=0.

The coefficient of t2y''(t)is a=1, the one multiplyingty'(t) is and the one multiplyingy(t) is c=17,and therefore the associated characteristic equation to the given differential equation is:

role="math" localid="1655211520406" r2+(9-1)r+17=0r2+8r+17=0r=-8±64-682r=-4±i

02

Finding the general solution.

When the roots to the characteristic equation are complex,r=α±β¾± and if t>0then the linearly independent solutions arey1(t)=tαcos(β±ô²Ô³Ù) andy2(t)=tαsin(β±ô²Ô³Ù). But if we have that t<0, then the linearly independent solutions are given as:

y1(t)=(-t)αcos(β±ô²Ô(-t) â¶Ä‰â¶Ä‰and â¶Ä‰â¶Ä‰y2(t)=(-t)αsin(β±ô²Ô(-t))

Hence, the general solution to the given differential equation is:

y(t)=c1(-t)-4cos(ln(-t))+c2(-t)-4sin(ln(-t))y(t)=c1t-4cos(ln(-t))+c2t-4sin(ln(-t))

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