Solve the initial value problem x^2y'' - 3xy' = (6y^2)/(x^2)…

Solve the initial value problem $x^2y'' - 3xy' = \frac{6y^2}{x^2} - 4y$, with $y(1)=1, \qquad y'(1)=4$.

Step 1: Rearranging

Move all terms to one side: $x^2y'' - 3xy' + 4y = \frac{6y^2}{x^2}.$

The left-hand side is exactly the (linear) Euler operator $x^2y''-3xy'+4y$; its characteristic equation $r^2-4r+4=(r-2)^2=0$ has the double root $r=2$, so $y=x^2$ is a resonant solution of the associated homogeneous Euler equation. This suggests the substitution $y=zx^2$, which factors out that resonance.

Step 2: The substitution $y=zx^2$

With $z=z(x)$ a new unknown function, \[y = zx^2, \qquad y' = z'x^2 + 2zx, \qquad y'' = z''x^2 + 4z'x + 2z.\]

Substituting into $x^2y''-3xy'+4y=6y^2/x^2$, the plain-$z$ terms arising from $x^2y''$, $-3xy'$, and $4y$ cancel completely — precisely because $x^2$ itself solves the homogeneous equation $x^2Y''-3xY'+4Y=0$ at the double root $r=2$, so $y=zx^2$ removes that resonant piece exactly. What remains is

\[ x^4z'' + x^3z' = 6z^2x^2. \]

Dividing by $x^2$: $x^2z'' + xz' = 6z^2.$

Step 3: The Euler substitution $x=e^t$

With $x=e^t$ (dot denotes $d/dt$), the standard Euler identities give $x^2z''=\ddot z - \dot z$ and $xz'=\dot z$, so

\[ (\ddot z - \dot z) + \dot z = 6z^2 \;\Longrightarrow\; \ddot z = 6z^2, \]

an autonomous second-order equation.

Step 4: An energy-type first integral

Multiply both sides by $\dot z$ and integrate: \[\dot z\,\ddot z = 6z^2\dot z \;\Longrightarrow\; \frac{d}{dt}\!\left(\frac{\dot z^2}{2}\right) = \frac{d}{dt}\bigl(2z^3\bigr) \;\Longrightarrow\; \frac{\dot z^2}{2} = 2z^3 + K.\]

Step 5: Transforming the initial conditions to $t=0$

At $x=1$, $t=\ln x=0$. Since $y=zx^2$, $z(0)=y(1)/1^2=1$. From $y'=z'x^2+2zx$, at $x=1$: $y'(1)=z'(1)+2z(1)=z'(1)+2=4$, so $z'(1)=2$. Since $\dot z=xz'$, we get $\dot z(0)=1\cdot z'(1)=2$.

Substituting $z(0)=1$, $\dot z(0)=2$ into the first integral: $\dfrac{4}{2}=2(1)^3+K \Rightarrow K=0$. So

\[ \dot z^2 = 4z^3. \]

Step 6: Solving the first-order equation for $z(t)$

Since $\dot z(0)=2>0$, take the positive root: $\dot z = 2z^{3/2}$. Separating variables: \[\frac{dz}{z^{3/2}} = 2\,dt \;\Longrightarrow\; -2z^{-1/2} = 2t + C_0.\]

Applying $z(0)=1$: $-2=C_0$, so

\[ -2z^{-1/2} = 2t - 2 \;\Longrightarrow\; z^{-1/2} = 1-t \;\Longrightarrow\; z = \frac{1}{(1-t)^2}. \]

Step 7: Returning to $x$, then to $y$

Since $t=\ln x$: $z = \frac{1}{(1-\ln x)^2}.$

Then $y=zx^2$ gives $y = \frac{x^2$, valid for $x>0$, $x\neq e$.

Step 8: Verification

At $x=1$: $y(1)=1/(1-0)^2=1. $\checkmark$$ Differentiating $y=x^2(1-\ln x)^{-2}$ by the product/chain rule: \[y' = 2x(1-\ln x)^{-2} + x^2\cdot 2(1-\ln x)^{-3}\cdot\frac1x = \frac{2x(1-\ln x) + 2x}{(1-\ln x)^3} = \frac{2x(2-\ln x)}{(1-\ln x)^3}.\]

At $x=1$: $y'(1)=\dfrac{2(2)}{1}=4. $\checkmark$$ A full symbolic substitution confirms that $x^2y''-3xy'+4y=6y^2/x^2$ holds identically for this $y$ (the computation of $y''$ via the quotient/chain rule is algebraically involved but routine, and was independently verified).