can someone please help me solve #3 a bernoulli equation. please show as much cl
ID: 2881167 • Letter: C
Question
can someone please help me solve #3 a bernoulli equation. please show as much clear work as possible ! thank you !
The graph of this family is given in Figure 2.13. Since lution is obtained from (8): After eliminating the parameter, we find this latter y x2. One can readily see that this function is n See Figure 2.14 EXERCISES 2.6 Answers to odd-numbered problems begin on page A In Problems 1-6 solve the given Bernoulli equation. dy dy dy 4. x y(xy dy 6. 3(1 5. x In Problems 7-10 solve the given differential e indicated initial condition.Explanation / Answer
3) given dy/dx =y(xy3 -1)
dy/dx =(xy4 -y)
dy =(xy4 -y)dx
dy +ydx=xy4dx
(1/y4)dy +(1/y4)ydx=xdx
(1/y4)dy +(1/y3)dx=xdx
let (1/y3)= v
differentiate with respect to x on both sides .
(-3/y4)dy = dv
==>(1/y4)dy =(-1/3)dv
now
(1/y4)dy +(1/y3)dx=x dx
=> (-1/3)dv +vdx=x dx
=> dv -3vdx=-3x dx
integrating factor =e-3dx
integrating factor =e-3x
multiply on both sides by integrating factor
=> dve-3x -3ve-3xdx=-3xe-3x dx
=>(ve-3x)'=-3xe-3x dx
integrate on both sides
=>(ve-3x)'=-3xe-3x dx
=>ve-3x=-3xe-3x dx
integration by parts :u dv =uv -v du . u =-3x , dv =e-3x dx, du =-3dx , v =(-1/3)e-3x
=>ve-3x=-3x(-1/3)e-3x -(-1/3)e-3x(-3dx)
=>ve-3x=xe-3x -e-3xdx
=>ve-3x=xe-3x +(1/3)e-3x +c
=>v=x+(1/3) +ce3x
=>v=(1/3)(3x +1 +ce3x)
=> (1/y3)=(1/3)(3x +1 +ce3x)
=> y3=3/(1+3x+ce3x)
=> y=[3/(1+3x+ce3x)]1/3 is the solution
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