This question explores the family of differential equations dy/dx=sqrt(􏰐 1 +􏰏( a*x )+ 􏰏 (2 *y)) for various values of the parameter a.  

For the case a = 􏰐 0 find the analytical solution that passes through the point (0, 1) and verify that this is a solution to the differential equation. Use this solution to find the value of y correct to 4 decimal placeswhen x=􏰐1. 

In maple i did

y:=(1/2)*x^2+sqrt(3)*x+1:
diff(y,x)
                             
i got the answer x + sqrt(3)

as shown in the markscheme. please cluld anyone help how to get y before this step and what to do after.

dy/dx=sqrt(1+(a*x)+(2*y))

for the case a=1, y=1 and x=0 construct a program for the runge-kutta method of order 2 with formulae as follows where f(x,y)=dy/dx.

k_1=h*f(x_n,y_n)

k_2=h*f(x_n+h,y_n+k_1)

y_(n+1)=y_n+1/2(k_1+k_2).

After creating a program obtain value of y correct to 4 decimal places when x=1 for h=0.1 and h =0.05.

The question explores the family of differential equations dy/dx = sqrt(1+(a*x)+(2*y)) for various values of the parameter a.

This figure shows the tangent field in the case a=1.

sketch a tangent field in the case a=-2.

need the following diagram on maple:

How can i answer iv on Maple?

A family of curves has polar equation r=cos^n (theta/n), 0<=theta,n*pi, where n is a positive even integer.

Previously Using t = theta as the parameter and finding  a parametric form of the equation of the family of curves it was shown that 

dy/dx = (sin(t)sin(t/n)-cos(t)cos(t/n)) /( sin(t)cos(t/n)+cos(t)sin(t/n)).

Is it possible to show on Maple with a program that there are n+1 points where the tangent to the curve is paralell to the y axis?