Question

8. Solve the differential equation y"+y=sin(2) cos(x).

Answer 1

We can solve the given initial value problem by using method of variation of parameters method.

So the most general solution is   Y = Y_C + Y_P

Y_C = Homogeneous solutin

Y_P = Particular solution.

Now the given initil value problem is y’’ + y = sinx cosx

So let us take the homogeneous equation for solving

So y’’ + y = 0    -------------------(4)

Now let us find the characteristic equation for homogeneous equation by assuming the solution y = e^rx not equla to Zero.

So y’ = r e^rx

And y’’ = r²e^rx

so substitute in (1)

y’’ + y = 0

r²e^rx + e^rx = 0

e^rx (r²+1) = 0

e^rx is not equal to zero so r²+1 = 0

so r²+1 = 0

r²  = -1

r²  = i²

r₁ = +i and r₂ = -i  =>  0 +/- i   in the form of a +/- ib so a = 0 and b = 1

so when the characteristic equation roots are complex and imaginary then homogeneous solution is in the form as given below.

Yc = C₁e^ax cos(bx)+ C₂e^ax sin(bx)

Yc = C₁e^(0)x cos(x)+ C₂e^(0)x sin(x)

Yc = C₁ cos(x)+ C₂ sin(x)

So now we need to find the wronskian for particular solution

Yc = C₁y₁(x) + C₂y₂(x)

So y₁(x) = cosx and y₂(x) = sinx

y₁’(x) = - sinx and y₂’(x) = cosx

And  W(y₁ , y₂) = y₁(x) y₂’(x) - y₁’(x) y₂(x)

W(y₁ , y₂) = (cosx) (cosx) – (-sinx) (sinx)

W(y₁ , y₂) = (cos²x) + (sin²x)

W(y₁ , y₂) = 1

And non homogeneous term g(x) = sinx cosx

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