Answer:
The level curves F(t,z) = C for any constant C in the real numbers
where
[tex]F(t,z)=z^3t^2+e^{tz}-4t+2z[/tex]
Step-by-step explanation:
Let's call
[tex]M(t,z)=2tz^3+ze^{tz}-4[/tex]
[tex]N(t,z)=3t^2z^2+te^{tz}+2[/tex]
Then our differential equation can be written in the form
1) M(t,z)dt+N(t,z)dz = 0
To see that is an exact differential equation, we have to show that
2) [tex]\frac{\partial M}{\partial z}=\frac{\partial N}{\partial t}[/tex]
But
[tex]\frac{\partial M}{\partial z}=\frac{\partial (2tz^3+ze^{tz}-4)}{\partial z}=6tz^2+e^{tz}+zte^{tz}[/tex]
In this case we are considering t as a constant.
Similarly, now considering z as a constant, we obtain
[tex]\frac{\partial N}{\partial t}=\frac{\partial (3t^2z^2+te^{tz}+2)}{\partial t}=6tz^2+e^{tz}+zte^{tz}[/tex]
So, equation 2) holds and then, the differential equation 1) is exact.
Now, we know that there exists a function F(t,z) such that
3) [tex]\frac{\partial F}{\partial t}=M(t,z)[/tex]
AND
4) [tex]\frac{\partial F}{\partial z}=N(t,z)[/tex]
We have then,
[tex]\frac{\partial F}{\partial t}=2tz^3+ze^{tz}-4[/tex]
Integrating on both sides
[tex]F(t,z)=\int (2tz^3+ze^{tz}-4)dt=2z^3\int tdt+z\int e^{tz}dt-4\int dt+g(z)[/tex]
where g(z) is a function that does not depend on t
so,
[tex]F(t,z)=\frac{2z^3t^2}{2}+z\frac{e^{tz}}{z}-4t+g(z)=z^3t^2+e^{tz}-4t+g(z)[/tex]
Taking the derivative of F with respect to z, we get
[tex]\frac{\partial F}{\partial z}=3z^2t^2+te^{tz}+g'(z)[/tex]
Using equation 4)
[tex]3z^2t^2+te^{tz}+g'(z)=3z^2t^2+te^{tz}+2[/tex]
Hence
[tex]g'(z)=2\Rightarrow g(z)=2z[/tex]
The function F(t,z) we were looking for is then
[tex]F(t,z)=z^3t^2+e^{tz}-4t+2z[/tex]
The level curves of this function F and not the function F itself (which is a surface in the space) represent the solutions of the equation 1) given in an implicit form.
That is to say,
The solutions of equation 1) are the curves F(t,z) = C for any constant C in the real numbers.
Attached, there are represented several solutions (for c = 1, 5 and 10)
