Cauchy problem, $x' -\frac{x}{t}=t, x(1)=0$

Question

I cannot solve this Cauchy problem: $x'-\frac{x}{t}=t, x(1)=0$

$p(t)=-\frac{1}{t} \Rightarrow u(t)=e^{-\int\frac{1}{t}dt}=e^{-ln|t|}$

$x' \cdot e^{-ln|t|}-\frac{x}{t} e^{-ln|t|}=t\cdot e^{-ln|t|}$

$(xe^{-ln|t|})'=t\cdot e^{-ln|t|}$

$xe^{-ln|t|}=t+C$

$x(t)=1+\frac{C}{t}$

So, In short, I was doing this problem by multiplying each side with e taken to the power of integral of $\frac{1}{t}$ and I get the result as $x(t)=1+\frac{C}{t}.$ The answer is completely off, I should get $x=t^2-t, t>0.$

Answer 1

I would follow the procedure I learned in my ODE course and textbook.

1) Write out the general form of the first order linear ODE

$$\frac{dx}{dt}+P(t)x=Q(t)$$

You are given

$$\frac{dx}{dt}-\frac{x}{t}=t \implies P(t)=-\frac1t,~Q(t)=t$$

2) Find the integrating factor

$$\mu(t)=\text{exp}{\int P(t)dt}=\text{exp}\int-\frac1t dt=\text{exp}\big({-\ln|t|}\big)=\frac{1}{t}$$

3) Multiply your original equation by the integrating factor

$$\frac1t\frac{dx}{dt}-\frac{x}{t^2}=1 $$

4) Reduce to $\frac{d}{dt}$ on the LHS

$$\frac{d}{dt}(\frac1t x)=1$$

5) Integrate both sides with respect to $t$

$$\frac x t=t+C \implies x=t^2+Ct$$

6) Evaluate the initial condition of $x(1)=0$

$$x(1)=0\implies 0=1+c \implies C=-1$$

7) Solve for $x(t)$

$$x(t)=t^2-t$$

where $t>0$ follows directly from step 2. The integrating factor of $\frac{1}{t}$ is only defined for $t>0$.

Answer 2

You have figured out that the integrating factor is $t$

So define $y$ by $x \equiv ty$ write the DE in terms of $y$

$$ (ty)' - \frac {ty}t = ty' +y-y=t $$

So $y'=1 \implies y=t+C \implies x=t^2+Ct$

use initial condition to get $C=-1$

Answer 3

The integrating factor is given by $\displaystyle\exp\left(-\ln(t)\right) = \frac{1}{t}$. Consequently, one has

\begin{align*} x^{\prime} - \frac{x}{t} = t \Longleftrightarrow \frac{x^{\prime}}{t} - \frac{x}{t^{2}} = 1 \Longleftrightarrow \left(\frac{x}{t}\right)^{\prime} = 1 \Longleftrightarrow \frac{x}{t} = t + k \Longleftrightarrow x(t) = t^{2} + kt \end{align*}

Since $x(1) = 0$, we conclude that $k = -1$. Thus the final solution is $x(t) = t^{2} - t$ for $t > 0$.