Question

This is solving differential equations by sepeartion of variables

solve for y such that
dy+k(y-70)dx=0
and y(1)=140

show the steps
the problem part for me is I'm stuck trying to seperate them and making the y's go with the dy and everything else go with the dx


show all work
show steps, specifically
1. separating and grouping the y's with dy and x with dx (remember that
`\int\limits {c} \, dx =cx` where c is a constant
2. write the C for the constant
3. subsitute the point (0,140) to solve for C
4. solve for y once you know what C is


show work and answer if you know

Answer 1

`\mathrm dy+k(y-70)\,\mathrm dx=0`

Separating variables gives


`(\mathrm dy)/(y-70)=-k\,\mathrm dx`

Integrating both sides gives


`\displaystyle\int(\mathrm dy)/(y-70)=-k\int\mathrm dx`

`\ln|y-70|+C_1=-kx+C_2`

`\ln|y-70|=-kx+C`

Solving for
`y` gives


`e^(\ln|y-70|)=e^(-kx+C)`

`y-70=e^(-kx)e^C`

`y=Ce^(-kx)+70`

As
`y(1)=140`, you get


`140=Ce^(-k)+70`

`70=Ce^(-k)`

`C=70e^k`

so the particular solution to this ODE is


`y=70e^ke^(-kx)+70`

`y=70e^(k(1-x))+70`

Unless you have any more information, you're done.