Lessons 5-9 PDF

Verifying Solutions
Example:
Determine whether the given function a) y = sinx b) y = 4e-x and c) y = Cex
is a solution of the differential equation y” - y = 0.
Solution:
a) y = sinx ; y’ = cosx ; y” = -sinx it follows that:
( -sinx ) - ( sinx ) = 0 ; -2sinx ≠ 0, thus it is not a solution.
b) y = 4e-x ; y’ = -4e-x ; y” = 4e-x it follows that:

( 4e-x ) - ( 4e-x ) = 0 thereby it is a solution.
c) y = Cex ; y’ = Cex and y” = Cex it follows that:

( Cex ) - ( Cex ) = 0 is a solution for any value of C.
Exercises:
Verify the following function if it is a solution of the given differential equation:
1) y = Ce4x for y’ = 4y
2) y = e-2x for 3y’ + 5y = -e-2x
3) y = C1sinx - C2cosx for y” + y = 0
4) x2 + y2 = Cy for y’ = 2xy/( x2 - y2 )
5) y2 - 2lny = x2 for =
Finding a Particular Solution
Example:
For the differential equation xy’ - 3y = 0 , verify that y = Cx3 is a solution and find
the particular solution determined by the initial condition y = 2 when x = -3.
Solution:
y = Cx3 then y’ = 3Cx2 and xy’ - 3y = x ( 3Cx2 ) - 3 ( Cx3 ) =
Therefore it is a solution, furthermore, the initial condition y = 2 when x = -3 yields
y = Cx3 2 = C ( -3 )3 thus C = −
Thus conclude that the particular solution is:
𝟐
y = - x3
𝟐𝟕
Exercises:
Check if the equation is a solution of the given differential equation and find the particular
solution determined by the given initial conditions.
1) y = Ax2 + 2x for xy’ - 2y = -2x

y(1) = 4
2) y = 2C1ex + 3C2e-2x for y” + y’ - 2y = 0

y(0) = 2
y’ ( 0 ) = -1
3) y = Asin2x + Bcosx - 4x2 for y” + 4y + 16x2 + 8 = 0

y ( 0 ) = -2
y’ ( 0 ) = 4
Family of Curves
An equation involving a parameter, as well as one or both of the coordinates of a point in a

plane, may represent a family of curves.
Example:
1) Find the differential equations of a family of circles, each having its center on the line
y = x and each passing through the origin.
Center at ( h , k )
but y = x
thus h = k = c
General equation of a circle:

( x - h )2 + ( y - k )2 = r2 but h = k and r2 = h2 + k2 = c2 + c2 = 2r2
( x - c )2 + ( y - c )2 = 2c2 ; x2 - 2xc + c2 + y2 - 2yc + c2 = 2c2
x2 + y2 - 2c ( x + y ) = 0
Differentiating once with respect to x, we have ; 2x + 2yy’ - 2c ( 1 + y’ ) = 0
Eliminating the arbitrary constant c by substitution , we have:
2c = ; substitute to the derivatives, we find that:
2x + 2yy’ - ( ) ( 1 + y’ ) = 0 will yield to:
𝒙𝟐 𝟐𝒙𝒚 𝒚𝟐
y’ =
𝒙𝟐 𝟐𝒙𝒚 𝒚𝟐
2) Find the differential equation satisfied by the family of parabolas having their vertices at
the origin and their foci on the y-axis.
Solution:
The general equation of a parabola opening upward or downward is:
( y - k ) = ±4a ( x - h )2
But since the vertices is at the origin in which ( h, k ) be ( 0 , 0 ) and foci on the y-axis and
±4a ( a as distance of vertex to foci ) be a constant A, then the equation becomes:
y = A x2 ( one arbitrary constant, derive once with respect to x )
y’ = 2Ax ; by substitution A = will yield to y’ = 2 ( ) x thus:
y’ = which gives xy’ - 2y = 0
3) Find the differential equation of the family of circles having their center’s on the y-axis.
Solution:
Since center is on y-axis, h = 0 thus equation be : x2 + ( y - h )2 = r2 ( 1 )
With two arbitrary constants , we will derive the equation twice with respect to x.
2x + 2 ( y - h ) y’ = 2r ; x + ( y - h ) y’ = r (2)
1 + ( y’ ) ( y’ ) + ( y - h ) y” = 0
( )
( y - h ) = - (3)
"
Substitute ( 2 ) and ( 3 ) in ( 1 ) will results to:

( ) ( )
x2 + [ - ]2 = [ x +( - ) y’ ] 2
" "
Simplifying :
( ) ( ) ( ) ( ) ( )
x2 + ( " )
= x2 - 2x ( "
) y’ + ( " )
( y’ )2
Final equation will yields to:

2xy”y’ [ 1 + ( y’ )2 ] - [ ( y’ )2 - 1 ] [ 1 + 2 ( y’ )2 + ( y’ )4 ] = 0
4) Find the differential equation of all family of straight lines with slope and y-intercept
equal.
Solution:
From the equation of a straight line in intercept form, we have:
y = mx + b but m = b yielding to y = mx + m = m( x + 1 )
Differentiating once with respect to x:
y’ = m ; substitute to the original equation y = y’ ( x + 1 )
Final equation be: y’ ( x + 1 ) - y = 0
Exercises:
1) Find the differential equation of all family of straight lines at a fixed distance p from the
origin.
2) Find the differential equation of family of circles tangent to the x-axis.
3) Find the differential equation of family of parabolas with vertex on the x-axis, with axis
parallel to the y-axis and with distance from focus to vertex fixed as a.
4) Find the differential equation of the family of parabolas with axis parallel to the x-axis.
5) Find the differential equation of the cubics : cy2 = x2 ( x - a ) with a held fixed.
6) Find the differential equation of the quartics : c2y2 = x ( x - a )3 with c held as fixed.
7) Find the differential equation of family of circles through the intersection of the circle
x2 + y2 = 1 and the line y = x. Use the “ u + kv “ form; that is, the equation:
x2 + y2 - 1 + k ( y - x ) = 0

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Verifying Solutions

b) y = 4e-x ; y’ = -4e-x ; y” = 4e-x it follows that:

c) y = Cex ; y’ = Cex and y” = Cex it follows that:

Verify the following function if it is a solution of the given differential equation:

2) y = e-2x for 3y’ + 5y = -e-2x

3) y = C1sinx - C2cosx for y” + y = 0

y = Cx3 then y’ = 3Cx2 and xy’ - 3y = x ( 3Cx2 ) - 3 ( Cx3 ) =

Therefore it is a solution, furthermore, the initial condition y = 2 when x = -3 yields

Thus conclude that the particular solution is:

1) y = Ax2 + 2x for xy’ - 2y = -2x

2) y = 2C1ex + 3C2e-2x for y” + y’ - 2y = 0

3) y = Asin2x + Bcosx - 4x2 for y” + 4y + 16x2 + 8 = 0

An equation involving a parameter, as well as one or both of the coordinates of a point in a

General equation of a circle:

2c = ; substitute to the derivatives, we find that:

2x + 2yy’ - ( ) ( 1 + y’ ) = 0 will yield to:

y’ = 2Ax ; by substitution A = will yield to y’ = 2 ( ) x thus:

y’ = which gives xy’ - 2y = 0

Substitute ( 2 ) and ( 3 ) in ( 1 ) will results to:

Final equation will yields to:

2) Find the differential equation of family of circles tangent to the x-axis.

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