Question

Consider the differential equation

Answer 1

\[\begin{align*}\frac{dy}{dt}&=t^2y+1+y+t^2\\\\
&=t^2(y+1)+y+1\\\\
&=(t^2+1)(y+1)\\\\
\frac{dy}{y+1}&=(t^2+1)\,dt\end{align*}\]

Answer 2

I was able to get that but I was wondering about the other ones.

Answer 3

For part (b): A linear \(n\)th order ODE is homogeneous if it has the form
\[f_n(x)y^{(n)}+\cdots+f_1(x)y'+f_0(x)y=0\]
and nonhomogeneous if the RHS is some nonzero function of \(x\); call it \(f(x)\).

You're asked to first write the ODE in the general linear form, i.e.
\[f_1(x)y'+f_0(x)y=f(x)\]
Such an equation will have a nonhomogeneous solution as well as an associated homogeneous solution, due to the principle of superposition.

This equation can be written in the linear form as
\[\frac{dy}{dt}-(t^2+1)y=t^2+1\]
The associated homogeneous solution is the solution to the homogeneous counterpart to this ODE, i.e.
\[\frac{dy}{dt}-(t^2+1)y=0\]
which can be found by several methods. It's separable, and you also have the option of solving it as you would any general linear ODE (integrating factor, etc.).

For part (c): Equilibrium solutions are those that satisfy \(\dfrac{dy}{dt}=0\). In this case, you would solve for \(y\) by setting \(\dfrac{dy}{dt}\) equal to zero.

Part (d): I'm not explicitly familiar with the "extended linearity principle" (I might know the idea behind it, or by another name), so I would have to look that up.