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

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1For question one: y'+y=1 so you just have linear DE \(\Large e^{H(x)}= \LARGE e^{^{\LARGE \int dx}}\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1and then use the product rule backwards, as you should probably know to.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Ok, 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)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.13 is same as 2, but different variables.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1And 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.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Oh, 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/(y²9) over to dy/dt, and then integrate both sides with respect to t.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1For question 1, if you want: \(\large\color{black}{ \displaystyle y'=y1 }\) \(\large\color{black}{ \displaystyle \frac{ dy}{dx}=y1 }\) \(\large\color{black}{ \displaystyle \frac{1}{y1}\frac{ dy}{dx}=1 }\) integrate both sides with respect to x. (you will also then have to solve for y)

Agent_A
 one year ago
Best ResponseYou've already chosen the best response.0Thank You, @SolomonZelman!
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