Show that function is a solution to the linear DE

[Painguy]

Homework Statement

Show that for any linear equation of the form

$\frac{dy}{dx}$ + P(x)y = 0

if y(x) is a soltuion, then for any constant C the function Cy(x) is also a solution

Homework Equations

The Attempt at a Solution

(dy/dx) + P(x)y=0
∫P(x)y=∫0
y=-C/P(x)

C(dy/dx) + CP(x)y=0
C((dy/dx) + P(x)y)=0
∫P(x)y=∫0
y=-C/P(x)

This all seems a little to simple. Is what I did correct?

 

[Simon Bridge]

Yes. In fact what you did was more complicated than it needs to be: you don't need the integration.

 

[LCKurtz]

Homework Statement

Show that for any linear equation of the form

$\frac{dy}{dx}$ + P(x)y = 0

if y(x) is a soltuion, then for any constant C the function Cy(x) is also a solution

Homework Equations

The Attempt at a Solution

(dy/dx) + P(x)y=0
∫P(x)y=∫0
y=-C/P(x)

C(dy/dx) + CP(x)y=0
C((dy/dx) + P(x)y)=0
∫P(x)y=∫0
y=-C/P(x)

This all seems a little to simple. Is what I did correct?

Yes. In fact what you did was more complicated than it needs to be: you don't need the integration.

No, it isn't correct at all. That isn't how you solve this DE and you don't need to solve it anyway. To show that Cy satisfies the DE, plug it into the DE and verify it works. You will need to use the fact that y solves it.

 

[Simon Bridge]

Urgh - I didn't notice the mistakes in the integration etc.
My bad. Not looking carefully enough. i.e.

(dy/dx) + P(x)y=0
∫P(x)y=∫0
y=-C/P(x)

... the 2nd line does not follow from the 1st, and the 3rd does not follow from the 2nd.
Using proper notation on line 2 should have told you this.

It's besides the point for your question - you don't need to solve the DE.
Iif you still don't understand what went wrong later, then we'll have to address that.