Solving a Homogeneous Linear DE: dy/dx - 3y = 0

RD12

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I am asked to solve the Homogeneous Linear DE: dy/dx - 3y = 0

I am trying to solve this by using an integrating factor by using the following procedure:

d/dx[e^int(P(x) dx) * y] = e^int(P(x) dx) * f(x)
During the step-by-step solution of the DE the text makes the statement:

e^(-3x) * dy/dx - 3e^(-3x) * y = 0 is the same as d/dx[e^(-3x) * y] = 0
My question is why are these two equations the same?

"int" in the equation above is "integral"
 
I am asked to solve the Homogeneous Linear DE: dy/dx - 3y = 0

I am trying to solve this by using an integrating factor by using the following procedure:

d/dx[e^int(P(x) dx) * y] = e^int(P(x) dx) * f(x)
During the step-by-step solution of the DE the text makes the statement:

e^(-3x) * dy/dx - 3e^(-3x) * y = 0 is the same asd/dx[e^(-3x) * y] = 0
My question is why are these two equations the same?

"int" in the equation above is "integral"

d/dx [e^(-3x) * y] .......... using chain-rule

=
e^(-3x) *d/dx[y] + y *d/dx[e^(-3x)]

= e^(-3x) *d/dx[y] + y *[(-3) *e^(-3x)] ..... and so on......
 
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I would have said "product rule", \(\displaystyle \frac{duv}{dx}= u\frac{dv}{dx}+ v\frac{du}{dx}\), rather than "chain rule":
\(\displaystyle \frac{d(e^{-3x}y}{dx}\)\(\displaystyle = e^{-3x}\frac{dy}{dx}+ y\frac{de^{-3x}}{dx}\)\(\displaystyle = e^{-3x}\frac{dy}{dx}- 3e^{-3x}y\).


But what I really want to say is that this looks like the hardest possible way to approach this differential equation! \(\displaystyle \frac{dy}{dx}- 3y= 0\) is separable. It can be written \(\displaystyle \frac{dy}{dx}= 3y\) and then \(\displaystyle \frac{dy}{y}= 3dx\) and that is easy to integrate.
 
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