Respuesta :
Answer:
The general solution of given differential equation is [tex]\frac{1}{5}ln|5z+2|=e^{\sin t}+C[/tex].
Step-by-step explanation:
The given differential equation is
[tex]\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t[/tex]
Taking out common factors.
[tex]\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t[/tex]
Using variable separable method, we get
[tex]\frac{dz}{5z+2}=e^{\sin t}\cos t dt[/tex]
Integrate both sides.
[tex]\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt[/tex]
[tex]I_1=I_2[/tex] .... (1)
First solve LHS,
[tex]I_1=\int\frac{dz}{5z+2}[/tex]
Substitute [tex]5z+2=u[/tex]
[tex]5\frac{dz}{du}=1[/tex]
[tex]dz=\frac{1}{5}du[/tex]
[tex]I_1=\frac{1}{5}\int\frac{du}{u}[/tex]
[tex]I_1=\frac{1}{5}ln|u|+C_1[/tex]
[tex]I_1=\frac{1}{5}ln|5z+2|+C_1[/tex]
Now, solve RHS,
[tex]I_2=\int e^{\sin t}\cos t dt[/tex]
Substitute [tex]\sin t=v[/tex],
[tex]\cos t\frac{dt}{dv}=1[/tex]
[tex]\cos tdt=dv[/tex]
[tex]I_2=\int e^{v}dv[/tex]
[tex]I_2=e^{v}+C_2[/tex]
[tex]I_2=e^{\sin t}+C_2[/tex]
Subtitle the values of I₁ and I₂ in equation (1).
[tex]\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2[/tex]
[tex]\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1[/tex]
[tex]\frac{1}{5}ln|5z+2|=e^{\sin t}+C[/tex]
Therefore the general solution of given differential equation is [tex]\frac{1}{5}ln|5z+2|=e^{\sin t}+C[/tex].
