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Sparkling_Jules
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#1
Report Thread starter 14 years ago
#1
I could do the ones where y=vx

But how do you do these:

(x+y)dy/dx = x^2 + xy + x + 1; y=v-x

x^2d2y/dx2 +xdy/dx + y = 0; x=e^v

xd2y/dx2 - 2dy/dx + x + 0; dy/dx=v
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Gaz031
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(x+y)dy/dx = x^2 + xy + x + 1; y=v-x
y=v-x
dy/dx = dv/dx - 1
d^2y/dx^2 = d^2v/dx^2

(x+v-x)(dv/dx-1) = x^2 + x(v-x) + x + 1
v(dv/dx - 1) = vx + 1 + x
vdv/dx - v = vx + 1 + x
vdv/dx = vx + v + 1 + x
vdv/dx = (v+1)(x+1)
v/(v+1)dv/dx = (x+1)
INT [v/(v+1)] dv = INT(x+1) dx
Integrate, then use the substitution y=v-x to give v=x+y
The left hand side can be done via a substitution t = v+1. The right hand side is trivially easy.
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Sparkling_Jules
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#3
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Whoa you're quick!!! I would rep you but it won't let me! You'll have to settle for a hug instead, *huggle*
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Gaz031
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x^2d2y/dx2 +xdy/dx + y = 0; x=e^v
x^2 (d^2y/dx^2) + x(dy/dx) + y = 0

x=e^v, dx/dv = d^2x/dv^2 = e^v. dv/dx = 1/e^v
[Via the chain rule] dy/dx = (dy/dv).(dv/dx) = (dy/dv)(e^-v)
d^2y/dx^2 = d/dx(dy/dv.e^-v) = [(dy/dv).-e^v] + e^-v(d^y/dv^2)
Then you can make the substitutions to tidy things up and eventually solve. Check the above though, as it looks awful on a computer screen.
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Gaz031
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#5
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x.d2y/dx2 - 2.dy/dx + x + 0; dy/dx=v
dy/dx = v
d^y/dx^2 = dv/dx

x.dv/dx - 2v + x = 0
x.dv/dx - 2v = -x
dv/dx - 2v/x = -1
You should recognise this as a typical first order differential equation. Find the integrating factor and integrate, then substitute v=dy/dx and integrate again.

Edit:
Whoa you're quick!!! I would rep you but it won't let me! You'll have to settle for a hug instead, *huggle*
No worries. I haven't sat P2-P6 myself yet, so i need to keep this content in my mind.
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Sparkling_Jules
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#6
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I love you!!! Seriously, my teacher would've confused the hell outta me but that all makes sense!
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