# Prove that every member of the family functions y=(1+c*e^t)/(1-c*e^t) is a solution of the differential equation y'=(1/2)(y^2-1).

### 3 Answers | Add Yours

y= 1+ce^t / 1-ce^t

or y = u/v

==> y'= u'v-uv' / v^2

y'= (ce^t)(1-ce^t)-(-ce^t)(1+ce^t)/ (1-ce^t)^2

= ce^t -(c^2)e^2t +ce^t -(c^2)e^2t / (1-ce^t)^2

= 2ce^t -2(c^2)et^2t / (1-ce^t)^2

= 2ce^t(1-ce^t)/(1-ce^t)^2= 2ce^t/(1-ce^t)

==> y'= 2ce^t/(1-ce^t).....(1)

Now substitute with y'

y'=1/2(y^2-1)= 1/2[(1+ce^t)^2/(1-ce^t)^2]-1

= (1/2)(1+2ce^t +(c^2)e^2t)/(1-ce^t)^2 - 1

We will substitute 1 with (1-ce^t)^2/(a-ce^t)^2

==> y'= (1/2) (4ce^t)/(1-ce^t)^2

= (2ce^t)/(1-ce^t)^2.....(2)

From (1) and (2) we norice that y value is 0. Then y is a solution for the equation.

First, we'll differentiate the given expression y=(1+c*e^t)/(1-c*e^t). For this reason, we'll use the quotient rule.

y'=[(1+c*e^t)'*(1-c*e^t)-(1+c*e^t)*(1-c*e^t)']/(1-c*e^t)^2

y'=[(c*e^t)*(1-c*e^t)+(c*e^t)*(1+c*e^t)]/(1-c*e^t)^2

We'll open the brackets:

y'=[c*e^t-(c*e^t)^2+c*e^t+(c*e^t)^2]/(1-c*e^t)^2

After reducing similar terms, we'll get:

y'=(2c*e^t)/(1-c*e^t)^2

Now, we'll work the right side for the given result of y', by substituting y with it's expression.

y'=(1/2)(y^2-1)=(1/2)[(1+c*e^t)^2/(1-c*e^t)^2 - 1]

y'=(1/2){[(1+c*e^t)^2-(1-c*e^t)^2]/(1-c*e^t)^2}

y'=(1/2){[(1+c*e^t+1-c*e^t)(1+c*e^t-1+c*e^t)]/(1-c*e^t)^2}

y' = (1/2)[(2)(2c*e^t)/(1-c*e^t)^2]

y'=(2c*e^t)/(1-c*e^t)^2

We notice that we've obtained the same expression for y', so the given function is a solution of the differential equation, for ever value of c.

y = (1+ce^t)/(1-ce^t)...........(1) , is solution of y = (1/2)(y^2-1).

Solution:

Multiplying both sides of (1) by 1-ce^t, we get:

y(1-ce^t ) = 1+ce^t. ...........(2). Or

y -1 = c(y+1)e^t . Or

c = (y-1)/{(y+1)e^t}.............(3).

Differentiating both sides of (2), we get:

y(1-ce^t)' +y'*(1-ce^t) = (1+ce^t)'

y*(-ce^t) +y'(1-ce^t) = ce^t. Or

y' (1-ce^t) = ce^t+yce^t = c(1+y) e^t. Or

y' = c(1+y)e^t / (1-ce^t). Substititutting here the value of c from (3),

y' = {(y-1)/[(y+1)e^t]}[(1+y^t)e^t](1 - (y-1) e^t/[(y+1)e^t])

= (y-1){1-(y-1)/(y+1)]

=(y-1)(y+1)/{y+1-(y-1)}

= (y^2-1)/2 which proves the enunciation.