- Thread starter
- #1

#### alane1994

##### Active member

- Oct 16, 2012

- 126

I was given the problem,

\(y^{\prime\prime}+9y=9\sec^2(3t)\)

My work as follows, please let me know if this is correct and where to go from here. I have hit a roadblock of sorts.

\(y^{\prime\prime}+9y=9\sec^2(3t)\)

\(y^{\prime\prime}+9y=0\\

m^2+9=0\\

m=3\\

y=\cos(3t)+\sin(3t)\)

So,

\(y=c_1\cos(3t)+c_2\sin(3t)\)

Where,

\(y_1=\cos(3t)\\

y_2=\sin(3t)\)

\(Y(t)=-y_1\int \dfrac{y_2g(t)}{W(y_1,y_2)}dt+y_2\int \dfrac{y_1g(t)}{W(y_1,y_2)}\)

\(y_1=\cos(3t)~~~y_1^{\prime}=-3\sin(3t)\)

\(y_2=\sin(3t)~~~y_2^{\prime}=3\cos(3t)\)

\(W=y_1y_2^{\prime}-y_1^{\prime}y_2\)

Look ok so far? Any thoughts as to how I should proceed?

*"Find the general solution of the given differential equation."*\(y^{\prime\prime}+9y=9\sec^2(3t)\)

My work as follows, please let me know if this is correct and where to go from here. I have hit a roadblock of sorts.

\(y^{\prime\prime}+9y=9\sec^2(3t)\)

\(y^{\prime\prime}+9y=0\\

m^2+9=0\\

m=3\\

y=\cos(3t)+\sin(3t)\)

So,

\(y=c_1\cos(3t)+c_2\sin(3t)\)

Where,

\(y_1=\cos(3t)\\

y_2=\sin(3t)\)

\(Y(t)=-y_1\int \dfrac{y_2g(t)}{W(y_1,y_2)}dt+y_2\int \dfrac{y_1g(t)}{W(y_1,y_2)}\)

\(y_1=\cos(3t)~~~y_1^{\prime}=-3\sin(3t)\)

\(y_2=\sin(3t)~~~y_2^{\prime}=3\cos(3t)\)

\(W=y_1y_2^{\prime}-y_1^{\prime}y_2\)

Look ok so far? Any thoughts as to how I should proceed?

Last edited: