Question

ty''-(t+1)y'+y=t^2 y1=e^t y2=t+1
differential equations 2nd order equations with variable coefficients

Answer 1

try some general quadratic form

Answer 2

is this supposed to be a Cauchy euler equation?

Answer 3

not too sure of the names though :)

Answer 4

what are those \(y_1,y_2\)? and @rperez36 yes it appears to be so

Answer 5

yes

Answer 6

Actually no I'm still confused. Where did you get \(y_1\) from?

Answer 7

i think using laplace

Answer 8

they are givens with the problem

Answer 9

\[y_1=e^t \]
\[y_2=t+1\]

Answer 10

laplace transforms seem good but we dont have initial conditions

Answer 11

I'm guessing you want us first to verify those two solutions satisfy the homogeneous equation $$ty''-(t+1)y'+y=0$$For \(y_1=e^t\) we have \(y=y'=y''=e^t\) hence:$$te^t-(t+1)e^t+e^t=0\\te^t-te^t-e^t+e^t=0\\0=0$$

Answer 12

they both are solutions i think..then the general form must be a linear combination of the two

Answer 13

For \(y_2=t+1\) we have \(y=t+1,y'=1,y''=0\) hence:$$-(t+1)+t+1=0\\0=0$$so indeed \(y_1,y_2\) are two linearly independent solutions to our homogeneous equation

Answer 14

In Problems 37 through 40, use variation of parameters
to find a general solution to the differential equation
given that the functions y1 and y2 are linearly independent
solutions to the corresponding homogeneous equation for
Remember to put the equation in standard form

Answer 15

@him1618 indeed but I believe she wants a particular solution to the non-homogeneous equation

Answer 16

for t>0

Answer 17

weve got to have a set of initial conditions for a particular solution i believe?

Answer 18

@him1618 eh sorta but you're right that we're not interested in a specific solution instead a general one

Answer 19

i would recommend laplace with assumptions that y(0) and y'(0) are 0

Answer 20

we haven't gone over laplace :/

Answer 21

then i think mrs perez id recommend taking a standard quadratic form, ax^2 + bx + c and putting it in place of y, and then equating coefficients to determine a ,b, c

Answer 22

the solution is
\[C_1e^t+C_2(t+1)-t^2\]

Answer 23

you can try what i said and then linearly combine that with the 2 solutions given

Answer 24

you need to use variation of parameters to find the particular solution. The given solutions are the corresponding solutions (the solutions that substitute to equal zero); the method of variation of parameters will help find the nonhomogenous part (that equals t^2).

Variation of parameters indicates that the solution is of some form
\[y_{p}=u_1y_1+u_2y_2\]

Answer 25

id forgotten that one

Answer 26

that wouldnt account for the t^2

Answer 27

the only notes I have from this chapter are about second order Cauchy euler equations and every example he worked in class would =0. Just substituting in variables into the auxillary equation am^2+(b-a)m+c=0

Answer 28

yes it accounts for the t^2. In fact, the instructions typed above indicates the method to use

Answer 29

In Problems 37 through 40, use variation of parameters
to find a general solution to the differential equation
given that the functions y1 and y2 are linearly independent
solutions to the corresponding homogeneous equation for
Remember to put the equation in standard form

Answer 30

@him1618 it very clearly states to use variation of parameters not the Laplace transform nor the method of undetermined coefficients

Answer 31

i dont know the question..my bad then

Answer 32

Have you ever used variations of parameters @rperez36?

Answer 33

yeah it was the chapter before this so ill review the problem and try to get the solution

Answer 34

\[u_1'=\frac{ -y_{2}f(x) }{ W }; u_2'=\frac{ y_1f(x) }{ W}\]
where W is the Wronskian of the y_1 and y_2

Answer 35

isn't this what i need to do first??

Answer 36

you need to find the wronskian first.

Answer 37

what about putting it in standard form? i did find the wronskian i got -e^t

Answer 38

with the given y's anyway i thought that we needed the y's from the general solution first. ill give and example

Answer 39

i got something different for the Wronskian. The given y's solve the homogenous part of the equation.

Answer 40

if you plug in y=e^t into the differential equation the result is 0 not t^2

Answer 41

@druminjosh what did you find for the Wronskian?

Answer 42

http://prntscr.com/1i2qi4
that is one of our examples from variation of parameters chapter. the solution y1&y2 are from the complimentary solution...i think that is what it is called anyway

Answer 43

and we also used the Y_c cos and sin for the determinates

Answer 44

we're doing exactly what's done in the example. y_1 and y_2 given here ARE from the complimentary solution @rperez36

Answer 45

sorry its kinda hard to see

Answer 46

those big determinant bars in your example are the Wronskian

Answer 47

ok that's what i though and used them for determinate so for the equation @druminjosh gave before for -y1 and y2 we use the givens again so it would be like
-(t+1)t^2/W??

Answer 48

almost the equations i gave above are for u_1' and u_2'

Answer 49

i named it trash file so i'll know to throw it away later, not becaue your problem is trash :)

Answer 50

The wronskian is -te^t

Answer 51

lol ok and i made a dumb mistake i put t in bottom right instead of 1

Answer 52

no worries, i figured it was a silly mistake

Answer 53

ok one solution at a time. The general solution we are trying to find is of the form:

\[y=u_1y_1+u_2y_2\]

Answer 54

\[u_1'=\frac{ -y_2f(x) }{ W }\]

Answer 55

you'll need to use integration by parts to find u_1