solve the following exact ordinary differential equation:(2tz^3 + ze^(tz) - 4) dt + (3t^2z^2 + te^(tz) + 2) dz = 0^3 = to the power of 3

Accepted Solution

Answer:The level curves F(t,z) = C for any constant C in the real numberswhere[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 form1) M(t,z)dt+N(t,z)dz = 0To see that is an exact differential equation, we have to show that2) [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 that3) [tex]\frac{\partial F}{\partial t}=M(t,z)[/tex]  AND4) [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 tso,[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)