## Agent_A one year ago Ordinary Differential Equations Question: (see photo)

1. Agent_A

2. SolomonZelman

For question one: y'+y=1 so you just have linear DE $$\Large e^{H(x)}= \LARGE e^{^{\LARGE \int dx}}$$

3. SolomonZelman

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

4. SolomonZelman

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)

5. SolomonZelman

3 is same as 2, but different variables.

6. anonymous

help!! me plzz

7. SolomonZelman

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.

8. SolomonZelman

ok, bye

9. SolomonZelman

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.

10. SolomonZelman

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)

11. Agent_A

Thank You, @SolomonZelman!