y_{1}, y_{2}, y_{3} are particular solutions of y'+a(x)y=b(x), so the

berljivx8

berljivx8

Answered question

2022-01-20

y1,y2,y3 are particular solutions of y+a(x)y=b(x), so the function y2y3y3y1 is constant.

Answer & Explanation

Thomas Nickerson

Thomas Nickerson

Beginner2022-01-20Added 32 answers

Step 1
If y1 and y2 are both particular solutions to
y+a(x)y=b(x)
then y1y2 is a solution to the associated homogeneous differential equation
y+a(x)y=0
Indeed, we have
(y1y2)+a(x)(y1y2)=y1y2+a(x)y1a(x)y2
=(y1+a(x)y1)(y2+a(x)y2)
=b(x)b(x)=0
In your situation, you then have the quotient of two (nonzero) solutions to the homogeneous equation
y+a(x)y=0
This is separable, and a solution to this equation is of the form Aea(x)dx for some constant A. So what you have is a quotient of the form
A1ea(x)dxA2ea(x)dx=A1A2
Esta Hurtado

Esta Hurtado

Beginner2022-01-21Added 39 answers

Step 1
Just thought Id
RizerMix

RizerMix

Expert2022-01-27Added 656 answers

Differentiate the given quotient and replace \(y_{i})

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