Respuesta :
Answer with explanation:
The given non Homogeneous linear differential equation is:
y" +4 y'=3 Sin 2 x-------(1)
Put , u=y'
Differentiating once
u'=y"
Substituting the value of , y' and y" in equation (1)
⇒u' +4u =3 Sin 2x
This is a type of linear differential equation.
Integrating factor [tex]=e^{4t}[/tex]
Multiplying both sides of equation by Integrating factor
[tex]e^{4 x}(u'+4u)=e^{4x}3 \sin 2x\\\\ \text{Integrating both sides}\\\\ue^{4x}=\int {3 \sin 2x \times e^{4x}} \, dx \\\\ue^{4x}=\frac{3e^{4x}}{2^2+4^2}\times (4\sin 2x -2 \cos 2x)\\\\ue^{4x}=\frac{3e^{4x}}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}\\\\ \text{Using the formula of}\\\\\int{e^{ax}\sin bx } \, dx=\frac{e^{ax}}{a^2+b^2}\times (a \sin bx-b \cos bx)+C[/tex]
where C and [tex]C_{1}[/tex] are constant of integration.
Replacing , u by , y' in above equation we get the solution of above non homogeneous differential equation
[tex]y'(x)=\frac{3}{20}\times (4\sin 2x -2 \cos 2x)+C_{1}e^{-4 x}[/tex]
