Question

Ordinary Differential Equations Question: (see photo)

Answer 1

For question one:
y'+y=1

so you just have linear DE

\(\Large e^{H(x)}= \LARGE e^{^{\LARGE \int dx}}\)

Answer 2

and then use the product rule backwards, as you should probably know to.

Answer 3

Ok, for question 2

\(\large\color{black}{ \displaystyle \frac{dz}{dt} =z^2+4 }\)

integrate both sides with respect to z
(and then you will have to solve for z)

Answer 4

3 is same as 2, but different variables.

Answer 5

help!! me plzz

Answer 6

And then for 4.

\(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{t^2y+t^2+y+1} }\)

\(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{t^2(y+1)+y+1} }\)

\(\large\color{black}{ \displaystyle \frac{dy}{dt}=\frac{ 1}{(t^2+1)(y+1)} }\)

\(\large\color{black}{ \displaystyle (y+1)\frac{dy}{dt}=\frac{ 1}{t^2+1} }\)

integrate both sides with respect to t.

Answer 7

ok, bye

Answer 8

Oh, actually, it would be better for question 2:
1/(4+z²) dz/dt=1

then integrate with respect to t on both sides,
and then solve for z.
------------------
and bring the 1/(y²-9) over to dy/dt, and then integrate both sides with respect to t.

Answer 9

For question 1, if you want:

\(\large\color{black}{ \displaystyle y'=y-1 }\)

\(\large\color{black}{ \displaystyle \frac{ dy}{dx}=y-1 }\)

\(\large\color{black}{ \displaystyle \frac{1}{y-1}\frac{ dy}{dx}=1 }\)

integrate both sides with respect to x.
(you will also then have to solve for y)

Answer 10

Thank You, @SolomonZelman!