P4 - Differential equations - help!

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In the first question, is it k^(2y) or (k^2) * y ??

Ok, your auxilary equation is
m^2 + k^2 = 0
m = +- ik

So the the complete general solution is
y = Ae^(ikx) + Be^(-ikx)
y = Acos(kx) + Bsin(kx)

Sub in the initial conditions you have and you get
A = 1

dy/dx = k(Bcos(kx) - Asin(kx))
1 = kB
B = 1/k

y = cos(kx) + (sin(kx)/k

I think

Reply 3

m:)ckel

deianra

2. Show that the substitution y = z cos x, where z is a function of x, reduces the differential equation:

cos^2x(d^2y/dx^2) + 2 cos x sin x(dy/dx) + 2y = x cos^3 x

...to the differential equation:

(d^2z/dx^2) + z = x

Hence find y given that y = 0 and (dy/dx) = 1 at x = 0

oh dear, i remember doing this the other day, but here goes:

(i) y = zcosx (remember that z is a function of x)

dy/dx = (dz/dx)cosx - zsinx

d²y/dx² = (d²z/dx²)cosx - (dz/dx)sinx - [(dz/dx)sinx + zcosx]
d²y/dx² = (d²z/dx²)cosx - 2(dz/dx)sinx - zcosx

Substituting this into the differential equation:

cos²x[(d²z/dx²)cosx - 2(dz/dx)sinx - zcosx] + 2cosxsinx[(dz/dx)cosx - zsinx] + 2zcosx = xcos³x

Divide by cosx to get:

cosx[(d²z/dx²)cosx - 2(dz/dx)sinx - zcosx] + 2sinx[(dz/dx)cosx - zsinx] + 2z = xcos²x

Now expand:

cos²x(d²z/dx²) - 2cosxsinx(dz/dx) - zcos²x + 2cosxsinx(dz/dx) - 2zsin²x + 2z = xcos²x

notice the cancelling in blue.

Now change the equation so that it is in terms of cos:

cos²x(d²z/dx²) - zcos²x - 2z(1-cos²x) + 2z = xcos²x

Expanding and simplifying:

cos²x(d²z/dx²) + zcos²x = xcos²x

Finally, divide by cos²x, leaving:

d²z/dx² + z = x

===============================

(ii) Auxiliary equation is given by: m² + 1 = 0
=> m = ± i
=> z = Asinx + Bcosx

Try, z = kx
dz/dx = k, d²z/dx² = 0

Substituting into original equation:

0 + kx = x
k=1

=> z = Asinx + Bcosx + x

Substituting z into the y = zcosx equation, you get:

y = Asinxcosx + Bcos²x + xcosx
dy/dx = A[cos²x - sin²x] - 2Bcosxsinx + cosx - xsinx

x=0, y=0 => 0 = 0 + B
B=0

x=0, dy/dx = 1, => 1 = A - 0 + 1 - 0
A=0

=> y = xcosx

Woohooo! :biggrin:

Reply 4

JamesF

deianra

Ah, that's lovely! Thanks :smile:

I got confused when I subbed in the initial conditions and it told me A was something like 1/k but that was meant to be B and argh! Thanks :smile:

Cool, i wasnt sure if i was right, i only just learned how to solve them so i could answer ur questions :redface:

Anyway, for the second,
y = zcosx
dy/dx = cosx.dz/dx - zsinx
d2y/dx^2 = cosx(d^2z/dx^2) - 2sinx(dz/dx) -zcosx

Sub these into the equation you have and it reduces to
x = d^2z/dx^2 + z

So the auxilary equation is m^2 + 1 = 0,
z = Acosx + Bsinx

Because the LHS of the original equation is first degree, we can assume that the solution to the particular integral is
z = Cx + D
Therefore
dy/dx = C
d^2y/dx^2 = 0

So we have
x = d^2z/dx^2 + z
x = Cx + D
Equating coefficients we get
D = 0, C = 1 => z = x

Therefore the complete general solution is
z = Acosx + Bsinx + x

y = z.cosx
When x = 0, y = 0, so therefore 0 = z.1 = z, so
0 = Acosx + Bsinx + x = A

dy/dx = cosx.(dz/dx) - zsinx, but x = 0, so
dy/dx = dz/dx
Therefore
1 = dz/dx = Bcosx - Asinx + 1
B = 0

z = x
y = xcosx

Reply 5

JamesF

Mockel, you have
z = kx
d2y/dx^2 = 0 => k = 1, so then isnt the complete general solution
z = Asinx + Bcosx + x not 1 ??

Reply 6

m:)ckel

JamesF

Mockel, you have
y = kx
d2y/dx^2 = 0 => k = 1, so then isnt the complete general solution
z = Asinx + Bcosx + x not 1 ??

ah yes....thanks for spotting it- must be all the typing getting into my head! :redface:

Reply 7

m:)ckel

deianra

Thanks you two, that's absolutely fab! :biggrin:

Mockel, rep tomorrow and thank you :smile:

thank you very much....after james spotted my mistake, i corrected it and it's right :smile:

Reply 8

mathz

deianra

Hey guys,

(a) Show that the substitution v = xy transforms the differential equation:

x(d^2y/dx^2) + 2(1+2x)(dy/dx) + 4(1+x)y = 32e^(2x), where x ≠ 0

...into the differential equation:

(d^2v/dx^2) + 4(dv/dx) + 4v = 32e^(2x)

- Meg x

denote dy/dx by y' d2y/dx2 by y'':
y=v/x
y'=v'/x-v/x^2
y''=v''/x-v'/x^2-v'/x^2+2v/x^3
eqn becomes
x(v''/x-v'/x^2-v'/x^2+2v/x^3)+2(1+2x)(v'/x-v/x^2)+4(1+x)v/x=32e^(2x)

v''-2v'/x+2v/x^2+2v'/x-2v/x^2+4v'-4v/x+4v/x+4v= 32e^(2x)
v''+4v'+4v= 32e^(2x)

Reply 9

RobbieC

Perhaps when I do P4, I just wont bother with this chapter... It looks HORRIBLE.

Reply 10

mathz

deianra

(a) Show that the substitution v = xy transforms the differential equation:

x(d^2y/dx^2) + 2(1+2x)(dy/dx) + 4(1+x)y = 32e^(2x), where x ≠ 0

...into the differential equation:

(d^2v/dx^2) + 4(dv/dx) + 4v = 32e^(2x)

(b) Given that v = ae^(2x), where a is a constant, is a particular integral of this transformed equation, find a.

(c) Find the solution of the differential equation:

x(d^2y/dx^2) + 2(1+2x)(dy/dx) + 4(1+x)y = 32e^(2x)

...for which y = 2e^2 and (dy/dx) = 0 at x = 1

(d) Determine whether or not this solution remains finite as x --> infinity

Thanks again!

- Meg x

v'' + 4v' + 4v = 32e^(2x)

v = ae^(2x)
v'=2ae^(2x)
v''=4ae^(2x)
so 4a+8a+4a=32 which gives a=2
aux eqn (m+2)^2=0 so m=-2

solution v=(Ax+b)e^(-2x)+2e^(2x)

v = xy
so xy=(Ax+b)e^(-2x)+2e^(2x)
y = 2e^2 and (dy/dx) = 0 at x = 1
x=1 y=2e^2
2e^2=(A+B)e^(-2)+2e^(2)
so A+B=0 >>>>>>>>> 1
y+xdy/dx={A-2(Ax+B)}e^(-2x)+4e^(2x)
hence
2e^2={A-2(A+B)}e^(-2)+4e^(2)
using 1
-2e^2=Ae^(-2)
so A=-2e^4 and B=2e^4
as x tends to infinity solution tends to infinity since e^2x/x tends to infinity