MHB How do you rewrite the given equation as a function of u(x)?

[karush]

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

 

[MarkFL]

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

$$\mu(x)=\exp\left(\int \frac{1}{x}\,dx\right)=?$$

 

[karush]

$$\mu(x)=\exp\left(\int \frac{1}{x}\,dx\right)=?$$

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

ok hopefully

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

 

[MarkFL]

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

ok hopefully

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

Yes, your integrating factor is correct. (Yes)

 

[karush]

Yes, your integrating factor is correct. (Yes)

$\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$
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 ??

 

[MarkFL]

Your rewrite should be:

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

 

[karush]

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!