Question

(D^2+9) y=sec 3x solve linear differential equation using general method​

Answer 1

Answer: $y(x) = C_{1} cos 3x+C_{2} sin 3x+\frac{cos3xlog(cos3x)}{9} +\frac{xsin3x}{3}$

Step-by-step explanation:

As the differential equation is

$(D^{2}+9)y=sec3x,$ Where D = $\frac{d}{dx}$,

As by the Auxiliary equation is ($D^{2} +9$)=0

⇒ D=±3i

Hence, C. F is,

$y_{c} = C_{1}cos 3x +C_{2} Sin 3x$,

Now we take $y_{1} =cos 3x, y_{2}=sin3x$

w = $\left[\begin{array}{cc}y_{1} &y_{2}\\y'_{1}&y'_{2}\end{array}\right]$ =$\left[\begin{array}{cc}cos3x &sin 3x\\-3sin3x&3cos3x\end{array}\right] =3$

As particular integral is,

$y_{p} =-y_{1} \int\limits \frac{y_{2}X }{w} dx+y_{2} \int\limits\frac{y_{1} X}{w} dx$

$=-cos3x \int\limits \frac{sin3xsin3x}{3} dx+sin3x\int\limits \frac{cos3xsec3x}{3} dx\\=-\frac{cos3x}{3}\int\limits tan3xdx+\frac{sin3x}{3} \int\limits dx\\=\frac{cos3xlog(cos3x)}{9} +\frac{xsin3x}{3}$

As the general method $y(x)=y_{c} +y_{p}$

$y(x) = C_{1} cos 3x+C_{2} sin 3x+\frac{cos3xlog(cos3x)}{9} +\frac{xsin3x}{3}$

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