# What is the general solution of the differential equation? : #y''+4y=2sin2x#

##### 2 Answers

The general solution is

#### Explanation:

This is a second order linear, non-homogenous ODE.

The general solution can be written as

Find

Solve the caracteristic equation

The solution is

Find a particular solution of the form

Plugging those values in the ODE

The general solution is

# y(x) = Acos(2x)+Bsin(2x) -1/2xcos(2x) #

#### Explanation:

We have:

# y'' + 4y=2sin2x# ..... [A]

This is a second order non-Homogeneous Differentiation Equation. The standard approach is to find a solution,

**Complementary Function**

The homogeneous equation associated with [A] is

# y''+4y= 0 #

And it's associated Auxiliary equation is:

# m^2+4 = 0 #

Which has two pure imaginary solutions

Thus the solution of the homogeneous equation is:

# y_c = e^(0x){Acos(2x)+Bsin(2x)} #

# \ \ \ = Acos(2x)+Bsin(2x) #

**Particular Solution**

With this particular equation [A], a probable solution is of the form:

# y = acos(2x)+bsin(2x) #

Where

# y = axcos(2x)+bxsin(2x) #

Let us assume the above solution works, in which case be differentiating wrt

# y' \ \= (b-2ax)sin2x+(2bx+a)cos2x#

# y'' = -4(ax-b)cos2x-4(bx+a)sin2x #

Substituting into the initial Differential Equation

# {-4(ax-b)cos2x-4(bx+a)sin2x} + 4{axcos(2x)+bxsin(2x)} = 2sin2x #

Equating coefficients of

#cos(2x): 4b = 0 => b =0#

#sin(2x): -4a=2 => a=-1/2 #

And so we form the Particular solution:

# y_p = -1/2xcos(2x) #

**General Solution**

Which then leads to the GS of [A}

# y(x) = y_c + y_p #

# \ \ \ \ \ \ \ = Acos(2x)+Bsin(2x) -1/2xcos(2x) #