In Problems 31-40, solve the initial value problem.

[2 cos(2x+y)-x²] dx+[cos(2x+y) +e^y] dy=0 , y(1)=0

Definition of Initial Value Problem:

By an initial value problem for an n^th-order differential equation F (x,y, dy/dx, ..., d²y/dxⁿ)=0, this mean define a solution to the differential equation on an interval I that satisfies at x₀ the n initial conditions,

y(x₀)=y₀,

dy/dx (x₀)=y₁,

.

.

.

d^n-1y/dx^n-1 (x₀)=y_n-1,

Where x₀∈ I and y₀, y₁, ..., y_n-1 are given constants.

Exactness:

If ∂M/∂y = ∂N/∂x. Otherwise, the equation is not exact.

Formulae to be used:

∫ x^adx= x^a+1/a+1 +C

∫ e^ax dx = e^ax/a +C

∫ sin x dx= -cos x+C

∫ cos x dx= sin x+C

Given that,

[2 cos(2x+y)-x²] dx+[cos(2x+y) +e^y] dy=0 , y(1)=0 ….. (1)

Let us check whether the given equation is exact or not. Then,

M=2 cos(2x+y)-x²

N= cos(2x+y) +e^y

Differentiate the value of M and N.

∂M/∂y= -2 sin (2x+y)

∂N/∂y= -2 sin (2x+y)

Therefore,

M dx + N dy=0

And

∂M/∂y - ∂N/∂y = -2 sin (2x+y) - [ -2 sin (2x+y) ]

= -2 sin (2x+y) +2 sin (2x+y)

=0

So, the given equation is exact.

Now, let us assume the following.

M=∂F/∂x

=2 cos (2x+y)-x²

Integrate on both sides, and you have

F=∫ (2 cos (2x+y)-x²) dy

= sin (2x+y) - x³/3 +g(y)

Differentiate the F with respect to y.

∂F/∂y = cos (2x+y) +g'(y)

=N

Equalize the values of N.

cos (2x+y) +g'(y) = cos(2x+y)+ e^y

g'(y) =e^y

Integrate on both sides.

∫ g'(y) = ∫e^y dy

g(y) = e^y +C₁

Substitute in the equation of F.

sin (2x+y)-x³/3+e^y+C₁=0

sin (2x+y)-x³/3+e^y=C ..... (2)

So, the solution is founded.

As given that,

y(1)=0

Then, x=1 and y=0.

Substitute the value in equation (2) to get the value of C.

sin (2x+y)-x³/3+e^y=C

sin (2+0) -1/3 +e⁰ =C

sin (2)+2/3=C

Substitute the value of C in equation (2).

sin (2x+y)-x³/3+e^y= sin (2)+2/3

sin (2x+y) =x³/3 - e^y+ 0.03489+0.66667

sin (2x+y) = x³/3 - e^y+ 0.70157

Hence, the solution of the given initial value problem is sin (2x+y) = x³/3 - e^y+ 0.70157.