MHB -b.2.2.1 de with u subst y' + (1/x)y=\sin x; x>0,

[karush]

665
$\textsf{Find the general solution of the given differential equation(book answer in red)}$
$$y^\prime + (1/x)y=\sin x \quad x>0, \qquad \color{red}
{\frac{c}{x}+\frac{\sin x}{x}-\cos x} $$
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]

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

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

Upon integrating, we get:

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

Hence:

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

 

[karush]

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

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

Upon integrating, we get:

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

Hence:

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

wait why would $u(x)=x$

 

[MarkFL]

wait why would $u(x)=x$

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