Find the general solution of x y'' + 2y' + xy = 0 , given…

Find the general solution of $x\,y'' + 2y' + xy = 0$, given that $y_1 = \dfrac{\sin x}{x}$ is a particular solution.

Step 1: Verify $y_1$

Differentiate $y_1 = \sin x / x$: \[y_1' = \frac{\cos x}{x} - \frac{\sin x}{x^2}, \qquad y_1'' = -\frac{\sin x}{x} - \frac{2\cos x}{x^2} + \frac{2\sin x}{x^3}.\]

Substituting into $xy''+2y'+xy$: \[x\!\left(-\frac{\sin x}{x}-\frac{2\cos x}{x^2}+\frac{2\sin x}{x^3}\right) + 2\!\left(\frac{\cos x}{x}-\frac{\sin x}{x^2}\right) + x\cdot\frac{\sin x}{x}\]

\[ = -\sin x - \frac{2\cos x}{x} + \frac{2\sin x}{x^2} + \frac{2\cos x}{x} - \frac{2\sin x}{x^2} + \sin x = 0.\ \checkmark \]

Step 2: Substitution $y = y_1 v$

Set $y = \dfrac{\sin x}{x}\,v(x)$. Then: \[y' = \frac{v'\sin x}{x} + v\!\left(\frac{\cos x}{x}-\frac{\sin x}{x^2}\right),\]

\[ y'' = \frac{v''\sin x}{x} + 2v'\!\left(\frac{\cos x}{x}-\frac{\sin x}{x^2}\right) + v\!\left(-\frac{\sin x}{x}-\frac{2\cos x}{x^2}+\frac{2\sin x}{x^3}\right). \]

Step 3: Reduce the order

Because $y_1$ solves the equation, every term multiplying $v$ alone reproduces Step 1's identity and cancels (times $v$), leaving only the $v'$, $v''$ terms: \[x\!\left[\frac{v''\sin x}{x} + 2v'\!\left(\frac{\cos x}{x}-\frac{\sin x}{x^2}\right)\right] + 2\cdot\frac{v'\sin x}{x} = 0,\]

\[ v''\sin x + 2v'\cos x - \frac{2v'\sin x}{x} + \frac{2v'\sin x}{x} = 0 \implies v''\sin x + 2v'\cos x = 0. \]

Step 4: Solve for $w=v'$

With $w=v'$, $w'=v''$, the equation $w'\sin x + 2w\cos x = 0$ is separable: \[\frac{dw}{w} = -2\cot x\,dx \implies \ln|w| = -2\ln|\sin x| + \mathrm{const} \implies w = \frac{A}{\sin^2 x}.\]

Step 5: Integrate for $v$

\[ v' = \frac{A}{\sin^2 x} \implies v = A\!\int \csc^2 x\,dx = -A\cot x + B. \]

Step 6: General solution

Then $y = \dfrac{\sin x}{x}(-A\cot x + B) = B\dfrac{\sin x}{x} - A\dfrac{\cos x}{x}$. Renaming $B=C_1$, $-A=C_2$:

Step 7: Verify $y_2 = \dfrac{\cos x}{x}$

\[ y_2' = -\frac{\sin x}{x} - \frac{\cos x}{x^2}, \qquad y_2'' = -\frac{\cos x}{x} + \frac{2\sin x}{x^2} + \frac{2\cos x}{x^3}. \]

\[ x y_2'' + 2y_2' + xy_2 = -\cos x + \frac{2\sin x}{x} + \frac{2\cos x}{x^2} - \frac{2\sin x}{x} - \frac{2\cos x}{x^2} + \cos x = 0.\ \checkmark \]

Step 8: Linear independence

\[ W(y_1,y_2) = y_1 y_2' - y_2 y_1' = -\frac{\sin^2x}{x^2} - \frac{\sin x\cos x}{x^3} - \frac{\cos^2x}{x^2} + \frac{\sin x \cos x}{x^3} = -\frac{1}{x^2} \neq 0, \] so $y_1,y_2$ form a fundamental system for $x \neq 0$.