It can be helpful to classify a differential equation, so that we can predict the techniques that might help us to find a function which solves the equation. Two classifications are the order of the equation -- (what is the highest number of derivatives involved) and whether or not the equation is linear .
Linearity is important because the structure of the the family of solutions to a linear equation is fairly simple. Linear equations can usually be solved completely and explicitly. Determine whether or not each equation is linear:
1. (1+y2)(d2y/dt2)+t(dy/dt)+y=et
2. t2(d2y/dt2)+t(dy/dt)+2y=sin t
3. (d3y/dt3)+t(dy/dt)+(cos2(t))y=t3
4. y''-y+y2=0

Respuesta :

Answer:

  1. Second-order nonlinear ordinary differential equation.
  2. Second-order linear ordinary differential equation.
  3. Third-order linear ordinary differential equation.
  4. Second-order nonlinear ordinary differential equation

Step-by-step explanation:

The objective is to determine whether or not each of the following equation is linear:

  1. [tex](1+y^2)\frac{d^2y}{dt^2} + t \frac{dy}{dt} = e^t[/tex]
  2. [tex]t^2\frac{d^2y}{dt^2} + t \frac{dy}{dt} +2y= \sin t[/tex]
  3. [tex]\frac{d^3y}{dt^3} + t \frac{dy}{dt} + \cos (2t) y = t^3[/tex]
  4. [tex]y''-y+y^2 = 0[/tex]

[tex](1)[/tex]

We can rewrite this equation in the form

                      [tex](1+y^2)y''(t) +ty'(t) = e^t[/tex].

As we can see, this is an second-order nonlinear differential equation, because of the term [tex]1+y^2[/tex] next to [tex]y''(t)[/tex].

[tex](2)[/tex]

We can rewrite this equation in the form

                         [tex]t^2y''(t) + t y'(t) +2y= \sin t[/tex].

This is an second-order linear ordinary differential equation.

[tex](3)[/tex]

We can rewrite this equation in the form

                         [tex]y'''(t)+ t y' + \cos (2t) y = t^3[/tex]

This is an third-order linear ordinary differential equation.

[tex](4)[/tex]

This is an second-order nonlinear ordinary differential equation, because of the term [tex]y^2[/tex].