## Creating Differential Equations Solution2

In this page creating differential equations solution2 we are going to see solutions of some practice questions.

Form a differential equations by eliminating arbitrary constants given in brackets against each.

(iii) x y = c²       {c}

Solution:

here we have only one arbitrary constant,so we can differentiate the given equation with respect to x only once.

x y = c²

x (dy/dx) + y (1) = 0

x y' + y = 0

Therefore the required equation is x y' + y = 0.

(iv) (x²/a²) + (y²/b²) = 1     {a , b}

Solution:

(x²/a²) + (y²/b²) = 1

(x²b² + y²a²)/a²b² = 1

x²b² + y²a² = a²b²

In this question we have two arbitrary constants. So we can differentiate the given equation two times.

differentiate the given equation with respect to x

2 x b² + 2 y y' a² = 0

divide the whole equation by 2

x b² + y y' a² = 0  ------ (1)

again differentiate the given equation with respect to x

we are going to differentiate y y' using product rule

u = y      v = y'

u' = y'     v' = y''

formula for product rule:

d (u v) = u v' + v u'

= y y'' + y' (y')

= y y'' + (y')²

(1) b² + [y y'' + (y')²] a² = 0

b² + [y y'' + (y')²] a² = 0 ----- (2)

=

 x y y' 1 y''+yy'

x (y'² + y y'') - y y' = 0

Therefore the required equation is  x (y'² + y y'') - y y' = 0.

(v) y = A e^(2x) + Be(-5x)   {A , B}

Solution:

Here we have two arbitrary constants. So we can differentiate the given equation two times.

y = A e^(2x) + Be(-5x)

y' = 2 A e^(2 x) - 5 Be^(-5 x)

y'' = 4 A e^(2 x) + 25 B e^(-5 x)

=

 y e2x e-5x y' 2e2x -5e-5x y'' 4e2x 25e-5x

taking e^(2x) and e^(-5x) commonly from second and third column,we get

=e2xe-5x

 y 1 1 y' 2 -5 y'' 4 25

e^(2x-5x) y(50 + 20) - y'(25 - 4) + y''(-5-2) = 0

e^(-3x) y(70) - y'(21) + y''(-7) = 0

70 y - 21 y' - 7 y'' = 0

now we are going to divide the whole equation by 7

10 y - 3 y' - y'' = 0

y'' + 3 y' - 10 y = 0

Therefore the required equation is y'' + 3 y' - 10 y = 0. 