Question 2.3.1: Solve the differential equation y'+(1+3t^2)y =0. In addition...

e will use integration to solve for y. Rearranging the given equation, we observe that y´=−(1+3t^{2})y. Dividing both sides by y, we find that

\frac {y´}{y}=−(1+3t^{2})

Keeping in mind the fact that y and y´ are each unknown functions of t, we integrate both sides of the previous equation with respect to t :

\int{\frac {y´}{y} dt} =\int{−(1+3t^{2}) dt}

Werecognize from the chain rule that the left-hand side is ln y. Thus, integrating the polynomial in t on the right yields

ln y =−t −t^{3} +C

Wenote that while an arbitrary constant arises on each side of the equation when integrating, it suffices to simply include one constant on the right. Finally, we solve for y using properties of the natural logarithm and exponential functions to find that

y = e^{−t−t^{3}+C} = e^{C} e^{−t−t^{3}}

Since C is a constant, so is e^{C}, and thus we write

y = Ke^{−t−t^{3}}

Observe that we have found an entire family of functions that solve the original differential equation: regardless of the constant K, the above function y is a solution. If we consider the stated initial-value problem and apply the given initial condition y(0) = 4, we immediately see that K = 4, and the solution to the initial-value problem is

y = 4e^{−t−t^{3}}