Bernoulli Differential Equations - Problems

1. Question 1
Solve
x y' +2 y= cos x
x
y' = dy
dx
x dy
dx +2 y = cos x
x
Rearranging
dy
dx + 2
x y= cosx
x2
Compare with Bernoulli differential
dy
dx + p y=Q
Here is the integrating factor
I . F=e ∫ pdx =e ∫ 2
x dx
=e2 ln x=eln x2
I . F=x2
Y ( I . F ) =∫ Q( I . F) dx
Y ( x2 ) =∫ cos x
x2 x2 dx
Y x2=sin x+ c
At x=π Y=0 give;
0 π2=sin π +c
C=0
So the final answer is Y x2=sin x

2. Question 2
Solve ( 1+ x6 ) y' = x2
y2 , y (0)=3
Solution
∫ y2 dy=∫ x2
( 1+ x6 ) + c
Let x3=t=x2 dx= dt
3
y3
3 = 1
3 tan−1
( ( 1+ x3 )
1 ) + c
y3=ta n−1 ( 1+ x3 )+ 3 c
Put x=0 making y=3 as stated in y (0)=3
33=tan−1 1+3 c
1
3 ( 27−ta n−1 )=c
c= 1
3 (−18 )=−6
Therefore
y3=ta n−1 ( 1+ x3 )−6

3. Question 3
Solve y' '+2 y' −8 y=x2 , y ( 0 )=0 , y' ( 0 )=0
This involves two steps
i. Get General Integral
ii. Get Partial solution
iii. Add step (i) and (ii)
Hence y=C . I + P . I
4.
Partial integral
y= x2
D2 +2 D−8
= −8 x2
1−D2+2 D
8
= −x2
8 (1− D2 +2 D
8 )
−1
=−x2
8 (1+ D2+ 2 D
8 +( D2 +2 D
8 )
2
)
=−x2
8 (1+ D2+ 2 D
8 + D4 +4 D2+ 4 D3
64 )
=−1
8 (x2 + 2+ 4 x
8 + 0+ 8+0
64 )
= −1
8 ( x2 + x
2 + 1
4 + 1
8 )
General integral
y ( D2 +2 D−8 ) =0
D2 +2 D−8=0
D=−2 ± √ 22 + 4 ×−8
2
D=−2 ± 6
2 =2 ,−4
y=C1 e2 x+C2 e−4 x at x=0 , y=0
C1+C2 ... ... ... ...i
dy
dx =2C1 e2 x+ 4 C2 e−4 x at x=0 , y '=0
0=2C1−4 C2
C1=2C2............ii
Solving i is (ii)
C2=0
D3 x2 →0
D . x2=2 x
Hence from the above solution; the differential equation; the differential equation
y=0+ (-1/8(x2 + x/2 + 3/8))
Hence y= -1/8(x2 + x/2 + 3/8))
Note:D: is the operator.D.x: Is the differentiation