Can some one answer these two questions

Mathematics

2 answers:

NeX [460]3 years ago

7 0

Answer:

1. Variable Term

2. Constant Term

Step-by-step explanation:

Alisiya [41]3 years ago

3 0

<h2>ANSWER:</h2>

1. Variable term

2.Constant term

first question is not clearly visible please mention it in comments

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Mice21 [21]

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4 years ago

(Differential Equations) Put these equations in explicit form:

Nuetrik [128]

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:

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