Given that y_1 = -(x)/(√((x^2+1)^3)) is a particular…

Given that $y_1 = -\dfrac{x}{\sqrt{(x^2+1)^3}}$ is a particular integral, find the general solution of $(x^2+1)y'' + 5xy' + 4y = 0$.

Step 1: Verifying $y_1$

Write $y_1=-x(x^2+1)^{-3/2}$. Differentiating, $y_1' = -(x^2+1)^{-3/2} + 3x^2(x^2+1)^{-5/2} = (2x^2-1)(x^2+1)^{-5/2}$.

\[ y_1'' = 4x(x^2+1)^{-5/2} - 5x(2x^2-1)(x^2+1)^{-7/2} = 3x(3-2x^2)(x^2+1)^{-7/2}. \]

Substituting into the left side (all terms share the factor $(x^2+1)^{-5/2}$): $(x^2+1)y_1''+5xy_1'+4y_1 = (x^2+1)^{-5/2}\,x\Bigl[3(3-2x^2)+5(2x^2-1)-4(x^2+1)\Bigr]$.

The bracket is $9-6x^2+10x^2-5-4x^2-4 = (9-5-4)+(-6+10-4)x^2 = 0$, so the equation is satisfied identically.\ $\checkmark$

Step 2: Abel's formula for the Wronskian

In standard form, $y''+\dfrac{5x}{x^2+1}y'+\dfrac{4}{x^2+1}y=0$, so $P(x)=5x/(x^2+1)$. Abel's formula gives

\[ W(x) = W_0\exp\!\left(-\int P\,dx\right) = W_0\exp\!\left(-\frac52\ln(x^2+1)\right) = \frac{W_0}{(x^2+1)^{5/2}}. \]

Step 3: Reduction of order

Seek a second solution $y_2=y_1v$. Since $W(y_1,y_2)=y_1y_2'-y_1'y_2=y_1^2v'$, taking $W_0=1$ in Abel's formula gives the separable equation

\[ y_1^2v' = \frac{1}{(x^2+1)^{5/2}}. \]

With $y_1^2 = x^2/(x^2+1)^3$, this becomes

\[ v' = \frac{(x^2+1)^3}{x^2(x^2+1)^{5/2}} = \frac{\sqrt{x^2+1}}{x^2}. \]

Step 4: Integrating $v'$ by parts

Take $u=\sqrt{x^2+1}$, $dw=x^{-2}dx$, so $du=\dfrac{x}{\sqrt{x^2+1}}dx$, $w=-\dfrac1x$: \[v = \int\frac{\sqrt{x^2+1}}{x^2}\,dx = -\frac{\sqrt{x^2+1}}{x} + \int\frac{dx}{\sqrt{x^2+1}} = -\frac{\sqrt{x^2+1}}{x} + \ln\bigl(x+\sqrt{x^2+1}\bigr) + C_2.\]

Dropping the constant $C_2$ (it only reproduces a multiple of $y_1$) keeps the genuinely new part: $v = -\frac{\sqrt{x^2+1}}{x} + \ln\bigl(x+\sqrt{x^2+1}\bigr)$.

Step 5: Assembling $y_2$

Multiply $v$ by $y_1=-x(x^2+1)^{-3/2}$ term by term. The first term collapses cleanly: \[\left(-\frac{x}{(x^2+1)^{3/2}}\right)\left(-\frac{\sqrt{x^2+1}}{x}\right) = \frac{\sqrt{x^2+1}}{(x^2+1)^{3/2}} = \frac{1}{x^2+1},\]

while the second term is left as is, so

Step 6: Verification: independence via the Wronskian

Because $y_2$ was built as $y_1v$, the Wronskian is automatically $W(y_1,y_2)=y_1^2v'$ — and Step 3 fixed $y_1^2v'$ to equal $1/(x^2+1)^{5/2}$ by construction. So, with no further computation needed,

\[ W(y_1,y_2) = \frac{1}{(x^2+1)^{5/2}} \neq 0 \quad\text{for all real } x, \]

which matches Abel's formula from Step 2 exactly (with $W_0=1$) and confirms $y_1,y_2$ are linearly independent, so $\{y_1,y_2\}$ is a genuine fundamental system.

Step 7: General solution

\[ y = C_1\frac{x}{(x^2+1)^{3/2}} + C_2\left[\frac{1}{x^2+1} - \frac{x\ln\bigl(x+\sqrt{x^2+1}\bigr)}{(x^2+1)^{3/2}}\right] \]

(the overall sign of the $y_1$ term has been absorbed into the arbitrary constant $C_1$).