Fp2 help needed . Differential equations

Hi guys .

Need some help with this . When doing questions how do I know when some of the particular integrals can’t be used . Like I know you can’t use something as a particular integral it’s part of the complementary function but in this case
https://postimg.cc/image/su7m0ocrb/

p isn’t in the CF here so why can’t we use it as the PI ? Why didn’t they use Pk otherwise instead of Pk^2

Reply 1

ghostwalker

Original post by Angels1234

H
p isn’t in the CF here so why can’t we use it as the PI ?

Huh! They did use p in the PI. See the line "Try y= p", then subbing that into the original equation produced the subsequent lines.

Reply 2

mqb2766

I'm not too sure of the question you're asking, but a few comments which may help.

Roughly
* Particular Integral (PI) basically looks at the right hand side of the equation
* Complementary function (CF) analyzes the left hand side
when you think of the right hand side as some forcing function, in this case a constant, the overall response is the sum of a transient (CF) and a steady state (PI).

The CF assumes the right hand side is zero and the exponential response (the differential equation is linear) is determined by the LHS and the initial conditions. You assume the response is y = exp(kx) and then substitute in to find two (2nd order) values for k. The overall response is
y = A*exp(k_1*x) + B*exp(k_2*x)
when the roots are repeated you have
y = A*exp(k*x) + B*x*exp(k*x)
as you have done correctly. A & B would be determined by the initial conditions y(0) and y'(0) - you've not given these (and to a small degree the PI).

The PI integral depends on the form of the RHS - in this case it is a constant. In this case, y must converge to a constant as t->inf (assuming k<0). Then as you say, the derivatives of a constant are zero, hence (again you've done this correctly) the solution must satisfy the equation and P = 1/(4k^2).

If the RHS was exponential, you'd assume the long term response would be exponential with the same exponent. Same for sinusoidal terms, if the rhs is sinusoidal, then y is sinusoidal. The response follows the right hand side after sufficiently large time, when the CF was decayed to zero.

The overall response is the sum of these two parts as the differential equation is linear.

Not easy, but you have to think of the CF and PI as two parts of the solution (at different times, CF when x is small, PI when x is large) and the overall solution is the sum of the two parts. As far as I can see, your attached solution is correct.

(edited 5 years ago)

Reply 3

Prasiortle

Original post by Angels1234

When the auxiliary equation has a repeated root, you use Ae^(kx) + Bxe^(kx), i.e. you go to x times the usual exponential term. Now when we put in y = p as the particular integral, this is just a constant, so its first and second derivatives are zero, so when substituted into the original differential equation, we just get k^2 * p = 1/4 (since the only term is the final k^2*y, and y = p), so p = 1/(4k^2).