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Differential Equations by Separation of Variables - Classwork dy = f ( x) > g ( y) . In order to solve it, you must put it in the form of dx g( y ) > dy = f ( x ) > dx allowing you to integrate. Your goal is to get an equation in the form of y = h( x)
A differential equation will be in the form of
dy
= y2 dx ! y #2dy =
dy 2 x
=
dx y 1)
!
y dy =
! 2 x dx
y = ± 2x2 + C
4)
!
!
dx
2) y #1
= x+C
#1
#1 y= x+C
y2
= x2 + C
2
dy
= 4y dx dy
! y=
dy x + sin( x )
=
3y 2 dx dy
= ky dx dy
! y=
! 4 dx
3)
3 y 2 dy =
y3 =
x2
# cos x + C
2
y=3
x2
# cos x + C
2
! k dx
5. ln y = kx + C
ln y = 4 x + C y = e 4 x +C
6)
y = e kx +C
y = Ce 4 x
! [ x + sin( x )] dx
dy
= xy dx dy
! y=
y = Ce kx
ln y = y =e
du
= e u+ 2t = e u > e 2t dt du
2t
7. ! u = ! e dt e 1
1
#e# u = e 2 t + C C u = # ln # e 2 t + C
2
2
! x dx
x2
+C
2
x2
+C
2
= Ce
x2
2
dx
= 1 + t # x # tx = (1 + t )(1 # x ) dt dx
= ! (1 + t ) dt
8. !
1# x t2 t2
+ t +C
+t
t2
# ln1 # x = t + + C 4 1 # x = e 2
4 x = 1 + Ce 2
2
Find the solution of the differential equation that satisfies the given condition. dx = 1 , x (0) = 1 dt x2 x dx = ! e t dt 4
= et + C
9. !
2
#1
1
= 1+ C 4 C =
4 x 2 = 2e t # 1
2
2
dy 1 + x
=
, y (1) = #4 dx xy
xe# t
dy
= y 2 + 1 , y (1) = 0 dx 1 dy = ! dx 4 tan#1 y = x + C
11. ! 2 y +1
1+ x y2 = ln x + x + C dx 4
10. ! y dy = ! x 2
8 = 1 + C 4 C = 7 4 y 2 = 2 ln x + 2 x + 14
x + 2y x2 + 1
12. 2 ! y dy =
y = tan( x + C ) 4 c + #1 4 y = tan( x # 1)
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AB Solutions
!
dy
= 0 , y (0) = 1 dx #x dx 4 y 2 = C # x 2 + 1
2
x +1
1 = C #1 4 C = 1 4 y2 = 2 # x2 + 1
- 212 -
Stu Schwartz
Differential Equations by Separation of Variables - Homework dy x
=
dx y
1.
! y dy = ! x dx 4
dy x 2 + 2
=
dx
3y 2
y2 x2
=
+C
2
2
!
A differential equation will be in the form of
dy
= y2 dx ! y #2dy =
dy 2 x
=
dx y 1)
!
y dy =
! 2 x dx
y = ± 2x2 + C
4)
!
!
dx
2) y #1
= x+C
#1
#1 y= x+C
y2
= x2 + C
2
dy
= 4y dx dy
! y=
dy x + sin( x )
=
3y 2 dx dy
= ky dx dy
! y=
! 4 dx
3)
3 y 2 dy =
y3 =
x2
# cos x + C
2
y=3
x2
# cos x + C
2
! k dx
5. ln y = kx + C
ln y = 4 x + C y = e 4 x +C
6)
y = e kx +C
y = Ce 4 x
! [ x + sin( x )] dx
dy
= xy dx dy
! y=
y = Ce kx
ln y = y =e
du
= e u+ 2t = e u > e 2t dt du
2t
7. ! u = ! e dt e 1
1
#e# u = e 2 t + C C u = # ln # e 2 t + C
2
2
! x dx
x2
+C
2
x2
+C
2
= Ce
x2
2
dx
= 1 + t # x # tx = (1 + t )(1 # x ) dt dx
= ! (1 + t ) dt
8. !
1# x t2 t2
+ t +C
+t
t2
# ln1 # x = t + + C 4 1 # x = e 2
4 x = 1 + Ce 2
2
Find the solution of the differential equation that satisfies the given condition. dx = 1 , x (0) = 1 dt x2 x dx = ! e t dt 4
= et + C
9. !
2
#1
1
= 1+ C 4 C =
4 x 2 = 2e t # 1
2
2
dy 1 + x
=
, y (1) = #4 dx xy
xe# t
dy
= y 2 + 1 , y (1) = 0 dx 1 dy = ! dx 4 tan#1 y = x + C
11. ! 2 y +1
1+ x y2 = ln x + x + C dx 4
10. ! y dy = ! x 2
8 = 1 + C 4 C = 7 4 y 2 = 2 ln x + 2 x + 14
x + 2y x2 + 1
12. 2 ! y dy =
y = tan( x + C ) 4 c + #1 4 y = tan( x # 1)
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AB Solutions
!
dy
= 0 , y (0) = 1 dx #x dx 4 y 2 = C # x 2 + 1
2
x +1
1 = C #1 4 C = 1 4 y2 = 2 # x2 + 1
- 212 -
Stu Schwartz
Differential Equations by Separation of Variables - Homework dy x
=
dx y
1.
! y dy = ! x dx 4
dy x 2 + 2
=
dx
3y 2
y2 x2
=
+C
2
2
!