To solve the differential equation y′′+4y=cos2(x)+sin2(x)tan2(x), we'll start by simplifying the right-hand side expression cos2(x)+sin2(x)tan2(x):
cos2(x)+sin2(x)=1 (because cos2(x)+sin2(x)=1 is the fundamental trigonometric identity)
So, we have:
cos2(x)+sin2(x)tan2(x)=1tan2(x)=tan2(x)
Now, the differential equation becomes:
y′′+4y=tan2(x)
This is a second-order linear homogeneous differential equation with a non-homogeneous term. To solve it, we'll first solve the homogeneous part y′′+4y=0, and then we'll find a particular solution for the non-homogeneous term tan2(x). Finally, we'll combine the solutions.
Homogeneous Part:
The characteristic equation for the homogeneous part is r2+4=0. Solving for r:
r2=−4
r=±2i
So, the characteristic roots are complex (imaginary). Therefore, the general solution to the homogeneous part is:
yh(x)=Acos(2x)+Bsin(2x)
Particular Solution:
To find a particular solution for the non-homogeneous term tan2(x), we'll use the method of undetermined coefficients. We assume a particular solution of the form:
yp(x)=Atan2(x)
Now, we'll find the derivatives needed for the equation:
yp′(x)=2Atan(x)sec2(x)
yp′′(x)=2A(sec2(x)+2tan2(x)sec2(x))=2Asec2(x)(1+2tan2(x))
Now, we can plug these derivatives into the original differential equation:
yp′′(x)+4yp(x)=2Asec2(x)(1+2tan2(x))+4Atan2(x)=2Asec2(x)(1+2tan2(x)+2tan2(x))=2Asec2(x)(1+4tan2(x))
Now, we equate this to the right-hand side of the original equation, which is tan2(x):
2Asec2(x)(1+4tan2(x))=tan2(x)
Simplifying:
2A(1+4tan2(x))=1
2A+8Atan2(x)=1
Now, we can equate coefficients:
2A=1⟹A=21
So, the particular solution is:
yp(x)=21tan2(x)
General Solution:
Now, we can combine the general solution to the homogeneous part and the particular solution: