Question

Find two solutions of the initial-value problem. (y'')²+ (y')²=1, y(π /2)= 1/2, y'(π /2)=√ 3/2

Use a numerical solver to graph the solution curves.

Short answer

The solution is y= 1- cos [x-(π /6)] , y= Sin [x-(π /3)]

Step-by-step solution

Solving for the given initial value problem;

Let suppose u = y’ so that u’ = y’’

Therefore, the DE becomes

(u') + (u)²=1

u'= ± √ 1- √ u²

Now, let us take solution u'= √ 1- √ (u)². Therefore we have

du/dx=√ 1- √ (u)²

du/[√ 1- √ (u)²] = dx

⟹ Sin^-1 u= x+c₁

sin^-1 y'= x+c₁

sin (x+c₁) = y'

Solving further:

At x= π /2, we have

Sin [π /2+c₁]=y'(π /2/)

Sin [π /2+c₁]= √ 3/2

Sin [(π /2)+c₁] = Sin (π /2)

c₁= - π /2

y'= Sin [x-(π/6)] +c₂

Again at x=π /2

y(π /2 )= -Cos[(π /2)-(π /6)] + c₂

1/2= -1/2+c₂

c₂=1

Therefore we have solution as

y= -cos[ x-(π /6) ]+1

y=1- cos [x-(π /6) ]

Drawing graph;

The graph for the solution is given in the figure

Solving for u'= - √ 1- √u2

Similarly, let take the solution u'= - √ 1- √u². Therefore we have

du/dx=- √ 1- √u²

-du/- √ 1- √u² =dx

⟹ Cos^-1 u= x+c₁

cos^-1 y'= x+c₁

cos^-1(x+c₁)= y'

At x= π /2 we have

cos[(π /2) +c₁]= y' (π /2 )

cos[(π /2) +c₁]= √3/2

cos[(π /2) +c₁]= cos (π /6)

c₁= -(π /3)

⟹ y= sin [x- (π /3)]

At x= π /2 , we have,

y (π /2 ) = Sin [π /2 -π /3] +c₂

1/2= 1/2+c₂

c₂=0

Therefore the solution isy= Sin [x-(π /3)]

The graph of solution is

The final answer isy= 1- cos [x-(π /6)] , y= Sin [x-(π /3)]