Find the solution of the following Differential Equations y′−4y=2x−4x2

Daniell Phillips

Daniell Phillips

Answered

2021-12-29

Find the solution of the following Differential Equations y4y=2x4x2

Answer & Explanation

censoratojk

censoratojk

Expert

2021-12-30Added 46 answers

Find the solution of the following DE
y4y=2x4x2
Complete solution =CF+PI
CF=y'-4y=0
(D4)y=0
m4=0
m=4
CF=C1e4x
PI=af(D)2x4x2D4
PI=14+D(2x4x2)
=14(1D4)(2x4x2)4+D=4(1D4)
=14(1D4)1(2x4x2)
(1D)1=1+D+D2+D3+
(1+D)1=1D+D2D3+
D(2x4x2)=ddx(2x4x2)=28x
D2(2x4x2)=d2dx2(2x4x2)=8
D3(2x4x2)=d3dx3(2x4x2)=0
PI=14(1+D4+(D4)2+(D4)3+)(2x4x2)

jean2098

jean2098

Expert

2021-12-31Added 38 answers

y4y=2x4x2 (1)
Clearly, the given differential equation is of the form y+py=a
Thus, differential eq (1) is linear, where P=4 and a=2x4x2
Now, integrating factor, I.F=epdx=edx
I.F=e4x
The solution of given differential equation (1) is given by,
y(IF)=q(IF)dx+c, c is a constant
y.e4x=(2x4x2)e4xdx+c (2)
Now (2x4x2)e4xdx=euu(u+2)16du
(2x4x2)edx4xeuu(u+2)du (3)
116euu(u+2)du=116[(u+2)(euueu)(euueu)du]
=116[(u+2)(euueu)euu+2eu]
=116[u2euueu+2ueu2euueu+2eu]
=116u2eu
Hence, from eq (3), we get
(2x4x2)e4xx=116u2eu
u=4x, we get
karton

karton

Expert

2022-01-09Added 439 answers

y.e4x=(2x4x2)e4xdx+c(2)(2x4x2)e4xdx=euu(u+2)16du(2x4x2)edx4xeuu(u+2)du(3)116euu(u+2)du=116[(u+2)(euueu)(euueu)du]=116[(u+2)(euueu)euu+2eu]=116[u2euueu+2ueu2euueu+2eu]=116u2euHence, from eq (3), we getint(2x4x2)e4xx=116u2euu=4x, we get(2x4x2)e4xdx=116×(4x)2e4x=x2e4xHence, from eq (2), we gety.e4x=x2e4x+cy=x2+ce4xAnswer: y=x2+ce4x

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