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Mathematics 8 Online
OpenStudy (anonymous):

Differential Equations SOS PLEASE

OpenStudy (anonymous):

\[y'''+y'=\frac{ sinx }{ \cos^2x }\]

OpenStudy (anonymous):

\[K^3+K=0\]

OpenStudy (anonymous):

\[K(K+1)=0\]

OpenStudy (anonymous):

meaning \[K1=0, K2,3= +-i\]

OpenStudy (anonymous):

\[Y=C1+C2\cos(x)+C3\sin(x)+Yp\]

OpenStudy (anonymous):

now how do I find Yp?

OpenStudy (anonymous):

should I find the derivative ?

ganeshie8 (ganeshie8):

familiar with variation of parameters ?

OpenStudy (anonymous):

yes, more or less

OpenStudy (anonymous):

@ganeshie8 ?

OpenStudy (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

OpenStudy (anonymous):

hmmmm not sure I'm following

OpenStudy (anonymous):

I found the general solution

OpenStudy (anonymous):

All I need now is the private one

OpenStudy (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 ?

OpenStudy (anonymous):

+C1

ganeshie8 (ganeshie8):

No, we're only looking at reduced DE for now.

OpenStudy (anonymous):

oh ok

OpenStudy (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)

OpenStudy (anonymous):

but what is g(t)?

OpenStudy (anonymous):

whatit refers to?

ganeshie8 (ganeshie8):

g(t) is whatever there on the right hand side

ganeshie8 (ganeshie8):

\(g(\color{red}{x}) = \dfrac{\sin x}{\cos^2x}\)

OpenStudy (anonymous):

and how do I find C1 then?

ganeshie8 (ganeshie8):

we will worry about that in the very end

OpenStudy (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 \(v\)

OpenStudy (anonymous):

roger that

OpenStudy (anonymous):

I think I can handle with that by myself, but how then can I find C1?

OpenStudy (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

OpenStudy (anonymous):

thank you so much @ganeshie8 !

OpenStudy (anonymous):

I think I've got it! I'll try at home and see if it works out for me

OpenStudy (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..

OpenStudy (usukidoll):

Y_h portion is always easier than finding the Y_p

OpenStudy (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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