Please Help Me I am a univercity student and cant understand :(

[asd103103]

Please solve:

1) solve using Maclauren serie y' = - xy : y( 0) = 1 : y' (0) = 0

2) D^2 y / dt^2 = x-2 and d^2 x / d t^2 = y+2

3) ( D^2 -1 ) y =x^2 e^x (I dont know exactly but it says chaging parameters.?)

4) Y' + y = 2 sin x -3 e^2 (solve this using the method of undetermined coefficents)

(The links show other places I've posted these questions.)

 

[stapel]

1) solve using Maclauren serie y' = - xy : y( 0) = 1 : y' (0) = 0

2) D^2 y / dt^2 = x-2 and d^2 x / d t^2 = y+2

3) ( D^2 -1 ) y =x^2 e^x (I dont know exactly but it says chaging parameters.?)

4) Y' + y = 2 sin x -3 e^2 (solve this using the method of undetermined coefficents)

Please reply showing your thoughts and efforts for each of these exercises, clearly showing where you are getting stuck. Thank you!

 

[HallsofIvy]

Please solve:

1) solve using Maclauren serie y' = - xy : y( 0) = 1 : y' (0) = 0

The second condition, y'(0) is redundant- if y'= -xy then y'(0)=-(0)(y(0))= -(0)(1)= 0.
But since this is a first order equation, the single condition is sufficient.
I presume you know that "MacLaurin series" for f(x) is \(\displaystyle y(0)+ y'(0)x+ y''(0)x^2+ ...\). You are told that y(0)= 1 and y'= -(0)(1)= 0. From y'= -xy, y''= -y- xy' so y''(0)= -y(0)- (0)(y'(0))= -1. Then y'''= -y'- y'- xy''= -2y'- xy'' so y'''(0)= -2(0)- 0(-1)= 0. Etc.

2) D^2 y / dt^2 = x-2 and d^2 x / d t^2 = y+2

Differentiating the first equation again, twice, \(\displaystyle d^4y/dt^4= d^2x/dt^2\) But the second equation says that \(\displaystyle d^2x/dt^2= y+ 2\), so \(\displaystyle d^4y/dt^4= y+ 2\). Can you solve that?

3) ( D^2 -1 ) y =x^2 e^x (I dont know exactly but it says chaging parameters.?)

Can you solve (D^2- 1)y= 0? I think you mean "variation of parameters". Do you know what that is?

4) Y' + y = 2 sin x -3 e^2 (solve this using the method of undetermined coefficents)

Can you solve y'+ y= 0? Was this actually "3e^2" or "3e^(2x)"? I am going to assume it was the latter. The "method of undermined coefficients" means that you put y= A x sin(x)+ B x cos(x)+ C e^{2x} into the equation and find A, B, and C such that the equation is satisfied. Add that to the general solution to y'+ y= 0.

(The links show other places I've posted these questions.)

I am puzzled by this. If, as you appear to be saying, you have never taken a "differential equations" course, where did you get these problems?