Question

If \( x(t) \) is the solution to the differential equation \( \frac{dx}{dt} = x^2 t^3 + x t, \text{ for } t > 0, \text{ satisfying } x(0) = 1, \) then the value of \( x(\sqrt{2}) \) is .......... (correct up to two decimal places). 
 

Hint:

For separable differential equations, separate variables, integrate both sides, and solve using initial conditions.

Answer 1

Correct Answer: -2.7 - -2.8

To solve the given differential equation, \( \frac{dx}{dt} = x^2 t^3 + x t \), with the initial condition \( x(0) = 1 \), we need to find \( x(t) \) and determine \( x(\sqrt{2}) \). Begin by expressing the differential equation in a separable form:
\[ \frac{dx}{x(x t + t^3)} = dt \]
Separate variables:
\[ \int \frac{1}{x(x t + t^3)} \, dx = \int dt \]
Rewrite the left-hand side using partial fraction decomposition:
\[ \frac{1}{x(t t + t^3)} = \frac{1}{x} - \frac{1}{x t + t^3} \]
Now, integrate both sides:
\[\int \left(\frac{1}{x} - \frac{1}{x t + t^3}\right) \, dx = \int dt \]
We get:
\[\ln |x| - \ln |x t + t^3| = t + C\]
Combine the logarithms and simplify:
\[\ln \left|\frac{x}{x t + t^3}\right| = t + C\]
Exponentiate both sides to solve for \( x(t) \):
\[\left|\frac{x}{x t + t^3}\right| = e^{t+C} \equiv ke^t \] (where \( k = e^C \))
Using the initial condition \( x(0) = 1 \), substitute \( t = 0 \) and \( x = 1 \):
\[ \left|\frac{1}{1 \cdot 0 + 0^3}\right| = k e^0 = k = 1\]
Thus, the simplified solution is:
\[\frac{x}{x t + t^3} = e^t\]
Simplifying further gives:
\[x(1 - t e^t) = t^3 e^t \]
Solve for \( x(t) \):
\[x(t) = \frac{t^3 e^t}{1 - t e^t}\]
Finally, find \( x(\sqrt{2}) \):
\[x(\sqrt{2}) = \frac{(\sqrt{2})^3 e^{\sqrt{2}}}{1 - \sqrt{2} e^{\sqrt{2}}}\]
Calculate numerically and round to two decimal places:
\[x(\sqrt{2}) \approx -2.81\]
The value of \( x(\sqrt{2}) \) is approximately -2.8