Question

Find the general solution of dz/dt=5ze^(sint)cost+2e^(sint)cost

Answer 1

Answer:

The general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$.

Step-by-step explanation:

The given differential equation is

$\frac{dz}{dt}=5ze^{\sin t}\cos t+2e^{\sin t}\cos t$

Taking out common factors.

$\frac{dz}{dt}=(5z+2)e^{\sin t}\cos t$

Using variable separable method, we get

$\frac{dz}{5z+2}=e^{\sin t}\cos t dt$

Integrate both sides.

$\int\frac{dz}{5z+2}=\int e^{\sin t}\cos t dt$

$I_1=I_2$         .... (1)

First solve LHS,

$I_1=\int\frac{dz}{5z+2}$

Substitute $5z+2=u$

$5\frac{dz}{du}=1$

$dz=\frac{1}{5}du$

$I_1=\frac{1}{5}\int\frac{du}{u}$

$I_1=\frac{1}{5}ln|u|+C_1$

$I_1=\frac{1}{5}ln|5z+2|+C_1$

Now, solve RHS,

$I_2=\int e^{\sin t}\cos t dt$

Substitute $\sin t=v$,

$\cos t\frac{dt}{dv}=1$

$\cos tdt=dv$

$I_2=\int e^{v}dv$

$I_2=e^{v}+C_2$

$I_2=e^{\sin t}+C_2$

Subtitle the values of I₁ and I₂ in equation (1).

$\frac{1}{5}ln|5z+2|+C_1=e^{\sin t}+C_2$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C_2-C_1$

$\frac{1}{5}ln|5z+2|=e^{\sin t}+C$

Therefore the general solution of given differential equation is $\frac{1}{5}ln|5z+2|=e^{\sin t}+C$.