The solution of the differential equation y′ + y/x = 18x + 4 can be written as kx² + hx − sy + C/x = 0. If s is a positive integer less than 3, find k, h and s.

Step-by-step solution

1. 1

   Identify the equation type and build the integrating factor. The equation is first-order linear in standard form $y'+P(x)y=Q(x)$ with $P(x)=1/x$ and $Q(x)=18x+4$. The integrating factor is

   $\mu(x)=e^{\int P\,dx}=e^{\int dx/x}=e^{\ln|x|}=x.$

   The whole point of $\mu$ is that multiplying by it makes the left side an exact derivative.

2. 2

   Multiply through and recognise the product rule.

   $xy'+y=18x^{2}+4x.$

   The left side is exactly $\dfrac{d}{dx}(xy)$ — this is the check that the integrating factor was computed correctly. So

   $(xy)'=18x^{2}+4x.$

3. 3

   Integrate both sides once.

   $xy=\int(18x^{2}+4x)\,dx=6x^{3}+2x^{2}+C,$

   and dividing by $x$,

   $y=6x^{2}+2x+\frac{C}{x}.$

4. 4

   Rearrange into the requested shape. Moving $y$ to the left,

   $6x^{2}+2x-y+\frac{C}{x}=0.$

   Comparing with $kx^{2}+hx-sy+C/x=0$ term by term forces $k=6$, $h=2$, $s=1$. The constraint "$s$ is a positive integer less than $3$" is what pins the normalisation down: the whole equation could be scaled by any constant, and only $s=1$ (rather than $s=2$, which would need $k=12$, $h=4$) leaves the $C/x$ coefficient equal to $1$ as written.

5. 5

   Verify by substituting back. With $y=6x^{2}+2x+Cx^{-1}$ we get $y'=12x+2-Cx^{-2}$, so

   $y'+\frac{y}{x}=12x+2-\frac{C}{x^{2}}+6x+2+\frac{C}{x^{2}}=18x+4.$

   The $C$ terms cancel for every $C$, confirming the general solution. Numerically, taking $C=1.7$ and $x=2.3$ gives $y'+y/x=45.4$, and $18(2.3)+4=45.4$.