# [SOLVED]2.2.1 de with u substitution

#### karush

##### Well-known member
$\textsf{Find the general solution of the given differential equation(book answer in red)}$
$$y^\prime + (1/x)y=\sin x \quad x>0, \qquad \color{red} {\frac{c}{x}+\frac{\sin x}{x}-\cos x}$$
ok first $u=1/x$ and $x=1/u$ then
$$u(x) = \exp\int u \, du = e^{\ln(u)}=u +c$$
proceed or ???

$\tiny{Elementary Differential Equations And Boundary Value Problems, \\ By: William E. Boyce and Richard C. Diprima \\ 1969, Second Edition}$

#### MarkFL

Staff member
We can see the integrating factor is $\mu(x)=x$ and so the ODE will become:

$$\displaystyle \frac{d}{dx}(xy)=x\sin(x)$$

Upon integrating, we get:

$$\displaystyle xy=\sin(x)-x\cos(x)+c_1$$

Hence:

$$\displaystyle y(x)=\frac{\sin(x)}{x}-\cos(x)+\frac{c_1}{x}$$

#### karush

##### Well-known member
We can see the integrating factor is $\mu(x)=x$ and so the ODE will become:

$$\displaystyle \frac{d}{dx}(xy)=x\sin(x)$$

Upon integrating, we get:

$$\displaystyle xy=\sin(x)-x\cos(x)+c_1$$

Hence:

$$\displaystyle y(x)=\frac{\sin(x)}{x}-\cos(x)+\frac{c_1}{x}$$
wait why would $u(x)=x$

#### MarkFL

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