Find that solution of y' = 2(2x-y) which passes through

Carole Yarbrough

Carole Yarbrough

Answered question

2021-12-26

Find that solution of y=2(2xy) which passes through the point (0, 1).

Answer & Explanation

Bukvald5z

Bukvald5z

Beginner2021-12-27Added 33 answers

Given y=2(2xy)(0,1)
dydx=2(2xy)
dydx+2y=4x
If =epdx=e2dx=e2x
y(IF)=(4x)(IF)dx+c
y(e2x)=4xe2xdx+c
y(e2x)=4[xe2xdx[ddx(x)e2xdx]dx+c
y(e2x)=4[xe2x2e2x2dx]+c
y(e2x)=4[xe2x2e2x4]+c
y(e2x)=44[2xe2xe2x]+c
y(e2x)=[2xe2xe2x]+c (a)
y(0)=1 from (a)
1(e0)=[2.0e0e0]+c
1=[01]+c
2=c
From (a)
y(e2x)=[2xe2xe2x]+2
Durst37

Durst37

Beginner2021-12-28Added 37 answers

Given: The differential equation y=2(2xy) and the point P(0,1).
Solution: y=2(2xy)
=4x2y
y+2y=4x
dydx+2y=4x
The above differential equation is in the form of first order linear differential equation.
dydx+P(x)y=Q(x)
Here P(x)=2 and Q(x)=4x
Integrating factor (IF)=eP(x)dx
=e2dx
=e2x
The general solution is
y×IF=(IF×Q(x))dx
y×e2x=(e2x×4x)dx
y×e2x=2e2xxe2x+C
y=2e2xxe2x+Ce2x passes through the point P(0,1)
1=2e2(0)(0)e2(0)+Ce2(0)

1=01+C1
1=1+C
C=2
The solution of the differential equation y=2(2xy) is
y(x)=2e2xxe2x+2e2x
y(x)=2x1+2e2x
user_27qwe

user_27qwe

Skilled2022-01-05Added 375 answers

y=2(2xy) and the point P(0,1).y=2(2xy)=4x2yy+2y=4xdydx+2y=4xdydx+P(x)y=Q(x)Here P(x)=2 and Q(x)=4x(IF)=eP(x)dx=e2dx=e2xThe general solution isy×IF=(IF×Q(x))dxy×e2x=(e2x×4x)dxy×e2x=2e2xxe2x+Cy=2e2xxe2x+Ce2xpasses through the point P(0,1)1=2e2(0)(0)e2(0)+Ce2(0)1=01+C11=1+CC=2y=2(2xy) isy(x)=2e2xxe2x+2e2xy(x)=2x1+2e2x

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