anonymous
  • anonymous
Differential Equations SOS PLEASE
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[y'''+y'=\frac{ sinx }{ \cos^2x }\]
anonymous
  • anonymous
\[K^3+K=0\]
anonymous
  • anonymous
\[K(K+1)=0\]

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anonymous
  • anonymous
meaning \[K1=0, K2,3= +-i\]
anonymous
  • anonymous
\[Y=C1+C2\cos(x)+C3\sin(x)+Yp\]
anonymous
  • anonymous
now how do I find Yp?
anonymous
  • anonymous
should I find the derivative ?
ganeshie8
  • ganeshie8
familiar with variation of parameters ?
anonymous
  • anonymous
yes, more or less
anonymous
  • anonymous
@ganeshie8 ?
anonymous
  • anonymous
I really need help with this one
ganeshie8
  • ganeshie8
\[y'''+y'=\frac{ \sin x }{ \cos^2x }\] Let \( y'=v\), then the DE becomes \[v''+v=\frac{ \sin x }{ \cos^2x }\] which is a familiar second order ordinary eqn, you can find \(Y_p\) using wrokskian or any other tricks that you're familiar with
anonymous
  • anonymous
hmmmm not sure I'm following
anonymous
  • anonymous
I found the general solution
anonymous
  • anonymous
All I need now is the private one
anonymous
  • anonymous
Are you familiar with the method I'm using?
ganeshie8
  • ganeshie8
http://gyazo.com/85b4d43f26ed36b347d32629ee0df5b0
ganeshie8
  • ganeshie8
\[v''+v=\frac{ \sin x }{ \cos^2x }\] earlier you worked the general solution, general solution for above reduced DE is \(c_2\cos x+c_3\sin x\), yes ?
anonymous
  • anonymous
+C1
ganeshie8
  • ganeshie8
No, we're only looking at reduced DE for now.
anonymous
  • anonymous
oh ok
anonymous
  • anonymous
Can you please show how to do it using the formula you've attached?
ganeshie8
  • ganeshie8
Okay, so from general solution we have \(y_1 = \cos x\) \(y_2 = \sin x\) Wronkskian\(W(y_1, y_2) = \begin{vmatrix} \cos x &\sin x\\\cos'x&\sin'x\end{vmatrix} = \cos^2x+\sin^2x = 1 \)
ganeshie8
  • ganeshie8
simply plug it in the formula and evaluate the integral(s)
anonymous
  • anonymous
but what is g(t)?
anonymous
  • anonymous
whatit refers to?
ganeshie8
  • ganeshie8
g(t) is whatever there on the right hand side
ganeshie8
  • ganeshie8
\(g(\color{red}{x}) = \dfrac{\sin x}{\cos^2x}\)
anonymous
  • anonymous
and how do I find C1 then?
ganeshie8
  • ganeshie8
we will worry about that in the very end
anonymous
  • anonymous
oh ok
ganeshie8
  • ganeshie8
Keep in mind we're solving the "reduced" DE in \(v\)'s completely first, then we're gona substitute back \(v\)
anonymous
  • anonymous
roger that
anonymous
  • anonymous
I think I can handle with that by myself, but how then can I find C1?
anonymous
  • anonymous
btw thanks a lot!
ganeshie8
  • ganeshie8
\(\large v(x) = c_2\cos x + c_3\sin x+Y_p\) plug in \(v(x) = y'\) back
ganeshie8
  • ganeshie8
and integrate
anonymous
  • anonymous
thank you so much @ganeshie8 !
anonymous
  • anonymous
I think I've got it! I'll try at home and see if it works out for me
ikram002p
  • ikram002p
good question, it was useful for me as well .
ganeshie8
  • ganeshie8
you may refer to this solution generated by wolfram if you get stuck..
1 Attachment
UsukiDoll
  • UsukiDoll
Y_h portion is always easier than finding the Y_p
UsukiDoll
  • UsukiDoll
there are 3 different ways to solve 2nd order odes... method of undetermined coefficients, variation of parameters, and laplace transform. It's just that sometimes one method is easier to use than the other 2. >_<

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