# [SOLVED]2.1.4 de with trig

#### karush

##### Well-known member
$$\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$$
first how do you get $u(x)$ from this

#### MarkFL

##### Administrator
Staff member
$$\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$$
first how do you get $u(x)$ from this
$$\displaystyle \mu(x)=\exp\left(\int \frac{1}{x}\,dx\right)=?$$

#### karush

##### Well-known member
$$\displaystyle \mu(x)=\exp\left(\int \frac{1}{x}\,dx\right)=?$$
$$\displaystyle\mu(x) =\exp\left(\int \frac{1}{x}\,dx\right)=e^{\ln{x}}=x$$

ok hopefully

ill funish this tomro
the latex is too hard to deal with on a tablet

#### MarkFL

##### Administrator
Staff member
$$\displaystyle\mu(x) =\exp\left(\int \frac{1}{x}\,dx\right)=e^{\ln{x}}=x$$

ok hopefully

ill funish this tomro
the latex is too hard to deal with on a tablet
Yes, your integrating factor is correct. #### karush

##### Well-known member
Yes, your integrating factor is correct. $\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$
multiply every term by x
$xy^\prime+y=3x\cos 2x$
rewrite as
$\displaystyle \frac{dy}{dx}=3x\cos 2x$
integrate
$\displaystyle y=\int 3x\cos 2x \, dx$

so far hopefully
the book answer is

$$\color{red} {\frac{c}{x} +\frac{3}{4}\frac{\cos 2x}{x} +\frac{3}{2}\sin 2x}$$

but ??

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#### MarkFL

##### Administrator
Staff member
Your rewrite should be:

$$\displaystyle \frac{d}{dx}(xy)=3x\cos(2x)$$

#### karush

##### Well-known member
rewrite as
$\displaystyle \frac{dy}{dx}(xy)=3x\cos 2x$
integrate
$$\displaystyle xy=\int 3x\cos 2x \, dx$$
then
$$\displaystyle xy=\frac{3}{2}x\sin(2x)+\frac{3}{4}\cos(2x)+c$$
divide by x
$$\displaystyle y=\frac{3}{2}\sin(2x)+\frac{3}{4}\frac{\cos(2x)}{x}+\frac{c}{x}$$

re-order and the book answer is:

$$\color{red} {\frac{c}{x} +\frac{3}{4}\frac{\cos 2x}{x} +\frac{3}{2}\sin 2x}$$

RAJ!!!

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