Special case when solving D.E.'s

[unseensoul]

Why can't there be common terms on both complementary function and particular integral when solving differential equations?

For instance,

dy/dx + 3y = exp(-x) + exp(-3x)

y(CF) = Aexp(−3x)

y(PI) = Cexp(−x) + Dxexp(−3x)

The term Dexp(-3x) in the P.I. has to be multiplied by x to be linearly independent of Aexp(-3x) in the C.F.. Why? What would happen if it wasn't?

 

[HallsofIvy]

Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.

 

[arildno]

Note that due to the characteristics of the differentiation of a product, ALL products of the form Y(x)=f(x)e^(-3x) will experience some "internal cancellation".

In your case, we will get on LHS:
$$\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}$$
Thus, since this must equal identically RHS, i.e, $$e^{-3x}$$, we get the subsidiary equation on f:
$$f^{,}(x)=1$$, which is readily solved.

 

[unseensoul]

Try it and see! Take y(x)= C exp(-x)+ D exp(-3x) and put it into the equation. Try to solve for C and D to make it true.

D would vanish... That's why they have to be linearly independent? I can't think of it in practical situations.

 

[unseensoul]

$$\frac{dY}{dx}+3Y=(f^{,}(x)e^{-3x}-3f(x)e^{-3x})+3f(x)e^{-3x}=f^{,}(x)e^{-3x}$$

Where did you get this from?

 

[arildno]

From defining Y(x) as f(x)e^(-3x), and applying the LHS operator on it!