SOLVE THE DIFFERENTIAL EQ AND CAUCHY PROBLEMS BELOW:
(Đã check bằngMathematica 9)
A. 1) y’ + 2xy = x
e
x
2

Sol: 𝐲
𝐞−𝐱
𝟐
𝐱
𝟐
𝟐+𝐂

2) (1 + x2)y’ – 2xy = (1 + x2)2
Sol: 𝐲
𝐱
𝟐+𝟏
𝐱+𝐂

3) y’ -
2y
x+1 = (x+1)3, y(0) =
1
2

Sol: 𝐲
𝐱+𝟏
𝟐
𝐱
𝟐
𝟐+𝐱+𝐂 C=1/2
4) (1 + x2)y’ + xy = 1, y(0) = 0
Sol: 𝐲
𝐂
𝐥𝐧
𝐱
𝐱
𝟐+𝟏
𝐱
𝟐+𝟏 C=0
5) xy’ – y = x2arctgx
Sol: 𝐲=𝐱
𝐂+𝐱
𝐭𝐚𝐧−𝟏
𝐱
𝐥𝐧
𝐱
𝟐+𝟏
𝟐

6) y’ -
y
xlnx = xlnx, y(e) =
e
2
2

Sol: 𝐲
𝐥𝐧
𝐱
𝐱
𝟐
𝟐+𝐂 C=0
7) (x3 + x)y’+ 3x2y =
x
2+1

Sol: 𝐲
𝐱
𝟐+𝟏
𝟑
𝟐
𝐥𝐧
𝐱
𝐱
𝟐
𝟐+𝐂

8) x(1+x2)y’ – (x2 - 1)y + 2x = 0
Sol: 𝐲
𝐱
𝟐+𝟏
𝐱
𝟏
𝐱
𝟐+𝟏+𝐂

9) y’ + xy = x3y3
Sol: 𝐲
𝐞
𝐱
𝟐
𝐂
𝐞−𝐱
𝟐
𝐱
𝟐+𝟏−𝟏
𝟐
𝟏
𝐱
𝟐+𝟏+𝐂
𝐞
𝐱
𝟐

10) (ylnx - 2)ydx = xdy
Sol: 𝐲
𝐱−𝟐
𝐂
𝟐
𝐥𝐧
𝐱+𝟏
𝟒
𝐱
𝟐−𝟏
𝟏
𝐂
𝐱
𝟐
𝐥𝐧
𝐱
𝟐
𝟏
𝟒

11) y’ + y =
e
x
2
y, y(0) =
9
4

Sol: 𝐲
𝐞−𝐱
𝐂
𝐞
𝐱
𝟐
𝟐 C=1
12) y’ -
y
2x = 5x2y5
Sol: 𝐲
𝐱
𝟏
𝟐
𝐂−𝟒
𝐱
𝟓−𝟏
𝟒

B.1) 2ydx + (y2 – 6x)dy = 0
2) ydx + (x + x2y)dy = 0
3)
dy
dx(x2y3 + xy) = 1
C. 1) (x + y + 1)dx + (x – y2 + 3)dy = 0
2) 2(3xy2 + 3y3)dx + 3(2x2y + y2)dy = 0
3) 3x2(1 + lny)dx – (2y -
x
3
y)dy = 0
4)
y
2(x−y
2
1
x
dx
1
y
x
2(x−y
2
dy=0
5)
1
y
sin
x
y
y
x
2
cos
y
x+1
dx
1
x
cos
y
x
x
y
2
sin
x
y
1
y
2
dy=0
D.Prove that theseequationshave a solutionisthequadraticformulaandsolveit:
1) x(x2 + 1)y’ – (2x2 + 3)y = 3
2) (x3 - x)y’ + (1 – 2x2)y + 1 = 0
E.Solve theequations byfindingthefactoranalysis:
1) (2xy + x2y +
y
3
3)dx + (x2 + y2)dy = 0; ((x)
2) y(1 + xy)dx – xdy = 0; ((y)
3) xdy + ydx – xy2lnxdx = 0; ((xy)
4) xdx + (2x + y)dy = 0; ((x + y)