Explanation:
To solve the given differential equation:
\frac{dx}{dt} = \frac{1+t}{tx}
We can rearrange the equation as:
\frac{x}{dx} = \frac{tx}{1+t} dt
Now we separate the variables by multiplying both sides by \frac{1+t}{x}:
\frac{x}{1+t} dx = t dt
Integrating both sides:
\int \frac{x}{1+t} dx = \int t dt
For the left side, we can use a substitution u = 1 + t, which gives du = dt:
\int \frac{x}{u} dx = \int (u - 1) du
Integrating further:
ln|x| = \frac{1}{2}u^2 - u + C
Substituting back u = 1 + t:
ln|x| = \frac{1}{2}(1+t)^2 - (1+t) + C
Simplifying:
ln|x| = \frac{1}{2}(1+t)^2 - (1+t) + C
ln|x| = \frac{1}{2}(1+t)^2 - 1 - t + C
Taking the exponential of both sides:
|x| = e^{\frac{1}{2}(1+t)^2 - 1 - t + C}
|x| = e^{\frac{1}{2}(1+t)^2 - 1 - t}e^C
Since e^C is just a constant, we can replace it with another constant A:
|x| = Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
Now we consider the absolute value of x. For x > 0:
x = Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
And for x < 0:
-x = Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
Combining these two cases, we can write the general solution as:
x = \pm Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
where A is an arbitrary constant.
We have obtained the general solution as:
x = ± Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
To simplify this expression, we can rewrite the exponent using algebraic manipulation:
x = ± Ae^{\frac{1}{2}(1+t)^2 - 1 - t}
= ± Ae^{\frac{1}{2}(1+t)^2 - t - 1}
= ± Ae^{\frac{1}{2}(1+t)^2 - 2t - 2}
= ± Ae^{\frac{1}{2}(1+t)^2} e^{-2t-2}
= ± Ae^{\frac{1}{2}(1+t)^2} e^{-2t} e^{-2}
Now, let's define a new constant B = ±Ae^{-2}. We can rewrite the solution as:
x = Be^{\frac{1}{2}(1+t)^2} e^{-2t}
So, the final solution to the given differential equation is:
x = Be^{\frac{1}{2}(1+t)^2} e^{-2t}
where B is an arbitrary constant.