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[SOLVED] 2.2.1 de with u substitution

karush

Well-known member
Jan 31, 2012
2,928
$\textsf{Find the general solution of the given differential equation(book answer in red)}$
$$y^\prime + (1/x)y=\sin x \quad x>0, \qquad \color{red}
{\frac{c}{x}+\frac{\sin x}{x}-\cos x} $$
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

Administrator
Staff member
Feb 24, 2012
13,775
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
Jan 31, 2012
2,928
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

Administrator
Staff member
Feb 24, 2012
13,775