Urgent help Homogeneous Linear D.E.

  • Thread starter rygza
  • Start date
  • #1
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Homework Statement


A linear equation in form:

dy/dx + P(x)y = 0 is said to be homogeneous since Q(x)=0.

a) show that y=0 is a trivial solution (wasn't even taught what a trivial solution is)
b) show that y=y1(x) is a solution and k is a constant, then y=ky1x is also a solution.
c) show that if y=y1x and y=y2x are solutions, then y=y1x + y2x is a solution

I don't even know how to start this problem. For part a) i simply plugged in 0 for y and got dy/dx=0 . doesn't seem right. then i tried separation of variable and got stuck at

(1/y)dy=-P(x)dx

can someone please guide me through? i have about three other problems like this and i haven't got a clue how to solve them.

p.s. that's y(sub1) and y(sub2)
 

Answers and Replies

  • #2
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you're not solving anything specific, since you don't really know about your functions. you're just proving general cases.

the trivial solution is just y=0. that's certainly true, dy/dx of 0 = 0, and P(x)y=P(x)0=0.

so, if you know that y1 is a solution, what do we know about constants? we can pull them out and apply them after we differentiate. so that's just like multiplying each term by k, and k0 = 0.

so we know that y1 and y2 are solutions in themselves. if you plug in y1 + y2 as a solution, well,
dy/dx (y1 + y2) = dy/dx y1 + dy/dx y2.
and P(x)(y1 + y2) = P(x)y1 + P(x) y2.
so dy/dx (y1 + y2) + P(x)(y1 + y2) = dy/dx y1 + dy/dx y2 + P(x)y1 + P(x) y2 = dy/dx y1 + P(x)y1 + dy/dx y2 + P(x)y2.
we know that dy/dx y1 + P(x)y1 = 0, and we know that dy/dx y2 + P(x)y2 = 0, since they're both solutions, so then dy/dx y1 + P(x)y1 + dy/dx y2 + P(x)y2 = 0.
i hope that's right, i'm a bit rusty with this stuff.
 

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