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Step 1: Rewrite the given differential equation in a more convenient form
The original differential equation is
$ \displaystyle \frac{dy}{dx} \;=\; \frac{y}{x}\,\Bigl(1 \;+\; x\,y^2\,(1 + \log_e x)\Bigr), \quad x>0, \quad y(1)=3.$
Rearrange this as
$ \displaystyle \frac{dy}{dx} \;-\; \frac{y}{x} \;=\; y^3\,(1 + \log_e x). $
Step 2: Transform the equation by dividing through by $y^3$
Divide every term by $y^3$ (assuming $y\neq 0$) to get
$ \displaystyle \frac{1}{y^3}\,\frac{dy}{dx} \;-\; \frac{1}{x\,y^2} \;=\; 1 + \log_e x. $
This suggests trying a substitution involving $y^{-2}$ because the left side has terms in $y^{-3}$ and $y^{-2}.$
Step 3: Substitution to simplify
Let $ \displaystyle t \;=\; -\frac{1}{y^2}. $
Then
$ \displaystyle \frac{dt}{dx} \;=\; -\,\bigl(-2\,y^{-3}\bigr)\,\frac{dy}{dx}
\;=\; \frac{2}{y^3}\,\frac{dy}{dx}. $
Hence
$ \displaystyle \frac{1}{y^3}\,\frac{dy}{dx}
\;=\; \frac{1}{2}\,\frac{dt}{dx}. $
Rewrite the differential equation in terms of $t$:
$ \displaystyle
\frac{1}{y^3}\,\frac{dy}{dx} \;-\; \frac{1}{x\,y^2}
\;=\; 1 \;+\; \log_e x
$
becomes
$ \displaystyle
\frac{1}{2}\,\frac{dt}{dx} \;-\; \frac{1}{x}\,\Bigl(-\,t\Bigr)
\;=\; 1 \;+\; \log_e x
$
(Notice $ y^{-2} = -t \implies -\frac{1}{x}\,y^{-2} = \frac{t}{x} .)$
Thus we get
$ \displaystyle
\frac{1}{2}\,\frac{dt}{dx} + \frac{t}{x}
= 1 + \log_e x.
$
Multiply through by 2 to simplify:
$ \displaystyle
\frac{dt}{dx} + \frac{2t}{x}
= 2\,\bigl(1 + \log_e x\bigr).
$
Step 4: Find the integrating factor
The above is a first-order linear differential equation in $t$:
$ \displaystyle
\frac{dt}{dx} + \frac{2}{x}\;t = 2\,(1 + \log_e x).
$
The integrating factor (I.F.) is given by
$ \displaystyle
\exp\!\Bigl(\int \frac{2}{x}\,dx\Bigr)
\;=\; \exp(2\log_e x)
\;=\; x^2.
$
Step 5: Multiply through by the integrating factor and integrate
Multiply both sides by $x^2$:
$ \displaystyle
x^2\,\frac{dt}{dx}
\;+\; 2\,x\,t
\;=\; 2\,x^2\,\bigl(1 + \log_e x\bigr).
$
Observe that the left-hand side is the derivative of $x^2\,t$:
$ \displaystyle
\frac{d}{dx}\bigl(x^2\,t\bigr)
\;=\; 2\,x^2\,(1 + \log_e x).
$
Integrate both sides with respect to $x$:
$ \displaystyle
x^2\,t
\;=\; 2 \int x^2\,\bigl(1 + \log_e x\bigr)\,dx \;+\; C,
$
where $C$ is the constant of integration.
Step 6: Evaluate the integral on the right
Compute
$ \displaystyle \int x^2(1 + \log_e x)\,dx
\;=\; \int x^2\,dx \;+\; \int x^2\,\log_e x\,dx.
$
$ \displaystyle \int x^2\,dx
= \frac{x^3}{3}.$
$ \displaystyle \int x^2\,\log_e x\,dx
= \frac{x^3}{3}\,\log_e x \;-\; \frac{x^3}{9}\quad
\bigl(\text{by integration by parts}\bigr).
$
Hence,
$ \displaystyle
\int x^2(1 + \log_e x)\,dx
= \frac{x^3}{3} \;+\; \Bigl(\frac{x^3}{3}\,\log_e x \;-\; \frac{x^3}{9}\Bigr)
= \frac{x^3}{3}\,\log_e x \;+\; \frac{2x^3}{9}.
$
Multiply by 2:
$ \displaystyle
2 \int x^2(1 + \log_e x)\,dx
= 2\Bigl(\frac{x^3}{3}\,\log_e x \;+\; \frac{2x^3}{9}\Bigr)
= \frac{2x^3}{3}\,\log_e x + \frac{4x^3}{9}.
$
Thus,
$ \displaystyle
x^2\,t
= \Bigl(\frac{2x^3}{3}\,\log_e x + \frac{4x^3}{9}\Bigr)
+ C
= \frac{2x^3}{9}\,\bigl(3\,\log_e x + 2\bigr) + C.
$
Step 7: Substitute back $ t = -\frac{1}{y^2}$
Recall $ t = -\tfrac{1}{y^2} $ so that
$ \displaystyle
x^2 \Bigl(-\frac{1}{y^2}\Bigr)
\;=\;
-\frac{x^2}{y^2}
= \frac{2x^3}{9}\,\bigl(3\,\log_e x + 2\bigr) + C.
$
Hence,
$ \displaystyle
\frac{x^2}{y^2}
= -\,\Bigl[\frac{2x^3}{9}\,(3\,\log_e x + 2) + C\Bigr].
$
Step 8: Use the initial condition $y(1) = 3$ to find $C$
When $x=1$, $y(1)=3$, so
$ \displaystyle
-\frac{1^2}{(3)^2}
= -\frac{1}{9}
= \frac{2\times 1^3}{9}\,\bigl(3\,\log_e 1 + 2\bigr) + C.
$
Since $\log_e 1 = 0,$ the above becomes
$ \displaystyle
-\frac{1}{9}
= \frac{2}{9}\,\bigl(2\bigr) + C
= \frac{4}{9} + C.
$
Thus
$ \displaystyle
C
= -\frac{1}{9} - \frac{4}{9}
= -\frac{5}{9}.
$
Step 9: Write the final solution for $y^2\!/9$
Substitute $C = -\tfrac{5}{9}$ back:
$ \displaystyle
-\frac{x^2}{y^2}
\;=\; \frac{2x^3}{9}\,(3\,\log_e x + 2) - \frac{5}{9}.
$
Move the negative sign:
$ \displaystyle
\frac{x^2}{y^2}
= \frac{5}{9} - \frac{2x^3}{9}\,\bigl(3\,\log_e x + 2\bigr).
$
Factor out $\tfrac{1}{9}$:
$ \displaystyle
\frac{x^2}{y^2}
= \frac{1}{9}\Bigl[\,5 \;-\; 2x^3\,(3\,\log_e x + 2)\Bigr].
$
Notice $ \log_e x^3 = 3\,\log_e x. $ Hence $(3\,\log_e x + 2)$ can be written as $(2 + \log_e x^3).$
Therefore,
$ \displaystyle
\frac{y^2}{9}
= \frac{x^2}{\bigl[5 - 2\,x^3\,\bigl(2 + \log_e(x^3)\bigr)\bigr]}.
$
This matches the stated correct answer:
$ \displaystyle
\frac{y^2(x)}{9}
= \frac{x^2}{5 \;-\; 2\,x^3\,\bigl(2 + \log_e x^3\bigr)}.
$
