Question

(Differential Equations) Put these equations in explicit form:

Answer 1

Answer:

1. $y= C(t*exp(-\frac{t^{2} }{2(2!)}+\frac{t^{4} }{4(4!)}-\frac{t^{6} }{6(6!)}+... ))}$

2. $t+C=ln(csc(y)-cot(y)$$23. [tex]Assuming t as independent variable:F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C\\Step-by-step explanation:1. Separable variables:[tex]\frac{dy}{dt}=\frac{y*cos(t) }{t}\\ \frac{dy}{y}= \frac{cos(t) }{t}dt\\ \int {\frac{dy }{y}} \, dt=\int {\frac{cos(t) }{t}} \, dt \\ln(y)-ln(C)=ln(t)-\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+... \\y=C(t*exp(\frac{t^{2} }{2(2!)} +\frac{t^{4} }{4(4!)} -\frac{t^{6} }{6(6!)}+...))$

2. Separable variables

\frac{dy}{sin(y)}=dt\\ \int\ \frac{1}{sin(y)}} \, dy = \int\ 1} \, dt\\t+C=ln(csc(y)-cot(y))[/tex]

3.  Homogeneous D.E

Rewriting:

$dr+(\frac{1}{m} exp(m+r)+r)dt=0\\\frac{dF}{dt}=1 -> F(r,t)=t+C(r)\\\frac{dF}{dy}=0+C'(r)= \frac{1}{m} exp(m+r)+r -> C(r)=\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} \\F(r,t)=t+\frac{1}{m} exp(m+r)+\frac{r^{2} }{2} =C$