Step 1

The given differential equation is,

\(\frac{dx}{dt}=1-t+x-tx\)

For solving the given differential equation using linear equation, first we separate the terms.

\(\frac{dx}{dt}=1-t+x-tx\)

\(\frac{dx}{dt}=(1+x)-t(1+x)\)

\(\frac{dx}{dt}=(1+x)(1-t)\)

\(\frac{dx}{(1+x)}=(1-t)dt\)

Step 2

Applying integration, on both the sides, we get

\(\int \frac{1}{(1+x)}dx=\int (1-t)dt\)

\(\ln(1+x)=t-\frac{t^{2}}{2}+C\)

\(1+x=e^{(t-\frac{t^{2}}{2}+C)}\)

\(x=e^{(t-\frac{t^{2}}{2}+C)}-1\)

Therefore, the general solution of the given differential equation is \(x=e^{(t-\frac{t^{2}}{2}+C)}-1\)

The given differential equation is,

\(\frac{dx}{dt}=1-t+x-tx\)

For solving the given differential equation using linear equation, first we separate the terms.

\(\frac{dx}{dt}=1-t+x-tx\)

\(\frac{dx}{dt}=(1+x)-t(1+x)\)

\(\frac{dx}{dt}=(1+x)(1-t)\)

\(\frac{dx}{(1+x)}=(1-t)dt\)

Step 2

Applying integration, on both the sides, we get

\(\int \frac{1}{(1+x)}dx=\int (1-t)dt\)

\(\ln(1+x)=t-\frac{t^{2}}{2}+C\)

\(1+x=e^{(t-\frac{t^{2}}{2}+C)}\)

\(x=e^{(t-\frac{t^{2}}{2}+C)}-1\)

Therefore, the general solution of the given differential equation is \(x=e^{(t-\frac{t^{2}}{2}+C)}-1\)