Give all solutions x(t) of (t - 1) x-dot - x + 1 = t for t > 1.

Step-by-step solution

1. 1

   Tidy the equation. Move the constant to the right: $(t-1)\dot{x}-x=t-1.$ Written this way the right-hand side is a clean multiple of $t-1$, which is the hint that the whole equation is about to collapse.

2. 2

   Normalise to standard linear form. Because $t>1$ we may divide by $t-1$: $\dot{x}-\frac{1}{t-1}x=1,$ which is $\dot{x}+p(t)x=q(t)$ with $p(t)=-\dfrac{1}{t-1}$ and $q(t)=1$.

3. 3

   Build the integrating factor. $\mu(t)=e^{\int p\,dt}=e^{-\ln(t-1)}=\frac{1}{t-1},$ where $\ln(t-1)$ needs no absolute value since $t-1>0$.

4. 4

   Multiply through and recognise the product rule. $\frac{\dot{x}}{t-1}-\frac{x}{(t-1)^2}=\frac{1}{t-1}\ \Longleftrightarrow\ \left(\frac{x}{t-1}\right)^{\!\prime}=\frac{1}{t-1}.$

5. 5

   Integrate and unwind. $\frac{x}{t-1}=\ln(t-1)+C\ \Longrightarrow\ x(t)=(t-1)\bigl(\ln(t-1)+C\bigr).$

6. 6

   Check by substitution. With $x=(t-1)(\ln(t-1)+C)$ we get $\dot{x}=\ln(t-1)+C+1$, so $(t-1)\dot{x}-x+1=(t-1)\bigl(\ln(t-1)+C+1\bigr)-(t-1)\bigl(\ln(t-1)+C\bigr)+1=(t-1)+1=t.$ The identity holds for every $C$, so this is the full family.