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**Hint:**Use the given solution \[{y_1} = \sin 3x\] to predict the required solution. Consider $y = a\sin x$ and find its single and double derivative then and use the given differential equation to simplify it. Then separate the variables and integrate them by substitution method and finally put the integrated value into the considered variable to get the solution.

**Complete step by step solution:**

We have given \[{y_1} = \sin 3x\] as one of the solution of the given differential equation $y'' + 9y = 0$, so we will assume $y = a\sin 3x$ to be its general solution, then

$

y' = a'\sin 3x + 3a\cos 3x\mathfrak{A} \\

{\text{And}}\; \\

y'' = a''\sin x + 3a'\cos x + 3a'\cos x - 9a\sin x \\

\Rightarrow y'' = a''\sin x + 6a'\cos x - 9a\sin x \\

$

From the given differential equation we know that,

$y'' + 9y = 0 \Rightarrow y'' = - 9y$

And also we have assumed $y = a\sin 3x$

$ \Rightarrow y'' = - 9\sin 3x$

Now putting this in the above equation, we will get

\[

\Rightarrow y'' = a''\sin x + 6a'\cos x - 9a\sin x \\

\Rightarrow - 9a\sin 3x = a''\sin x + 6a'\cos x - 9a\sin x \\

\Rightarrow a''\sin x + 6a'\cos x = 0 \\

\]

So we have reduced it to linear differential equation in terms of $a'$

We can also express the above expression as

\[ \Rightarrow \dfrac{{a''}}{{a'}} + \dfrac{{6\cos x}}{{\sin x}} = 0\]

After integrating both sides we will get

$ \Rightarrow \ln a' + 2\ln (\sin 3x) = \ln ( - 3{c_1})$

Using property of log, we can further write it as

$

\Rightarrow \ln a' + \ln ({\sin ^2}3x) = \ln ( - 3{c_1}) \\

\Rightarrow \ln (a'{\sin ^2}3x) = \ln ( - 3{c_1}) \\

$

Taking exponent to the base e both sides,

$

\Rightarrow {e^{\ln (a'{{\sin }^2}3x)}} = {e^{\ln ( - 3{c_1})}} \\

\Rightarrow a'{\sin ^2}3x = - 3{c_1} \\

\Rightarrow a' = \dfrac{{ - 3{c_1}}}{{{{\sin }^2}3x}} \\

$

Again integrating both sides, we will get

\[

\Rightarrow \int {a'} = \int {\dfrac{{ - 3{c_1}}}{{{{\sin }^2}3x}}} \\

\Rightarrow a = {c_1}\cot 3x + {c_2} \\

\]

So we have got the value of $a$

Putting this in $y = a\sin 3x$,

$

\Rightarrow y = ({c_1}\cot 3x + {c_2})\sin 3x \\

\Rightarrow y = {c_1}\cos 3x + {c_2}\sin 3x \\

$

**So this is the required solution for the differential equation $y'' + 9y = 0$**

**Note:**You can take anything for the constant part after integrating any function according to your use. As we have taken $\ln ( - 3{c_1})$ as the constant, because all the integrated terms in terms of log,\ so we have also taken the constant as log and also for further integration process we have taken $ - 3{c_1}$ so that the result will come out to be simplified and have less terms.

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