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#### karush

##### Well-known member

- Jan 31, 2012

- 2,928

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

first how do you get $u(x)$ from this

first how do you get $u(x)$ from this

- Thread starter karush
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- Thread starter
- #1

- Jan 31, 2012

- 2,928

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

first how do you get $u(x)$ from this

first how do you get $u(x)$ from this

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- #2

\(\displaystyle \mu(x)=\exp\left(\int \frac{1}{x}\,dx\right)=?\)$$\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$$

first how do you get $u(x)$ from this

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- #3

- Jan 31, 2012

- 2,928

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

=\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

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- #4

Yes, your integrating factor is correct.$$\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

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- #5

- Jan 31, 2012

- 2,928

$\displaystyle y^\prime +(1/x)y = 3\cos 2x, \quad x>0$Yes, your integrating factor is correct.

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 ??

Last edited:

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- Jan 31, 2012

- 2,928

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!!!

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

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!!!

Last edited: