Section 4.6

#4. The complementary solution to this DE is y=c1cos(x) + c2sin(x).
That tells us the particular solution is y = A cos(x) + B sin(x).
The two equations we need to solve simultaneously are: A' cos(x) + B' sin(x) = 0 and -A' sin(x) + B' cos(x) = sec(x) tan(x).
There are many ways to solve simultaneously. Whatever approach you choose, you should end up with:
A' = -tan2x = 1 - sec2x and B' = tan(x).
Integrating gives us: A = x - tan(x) and B = ln|sec(x)|
Thus our particular solution is: y = (x - tan(x))*cos(x) + sin(x)*ln|sec(x)| = x cos(x) - sin(x) + sin(x)*ln|sec(x)|
Since the "-sin(x)" can absorb into the complementary solution, our final solution to the DE is:
y = c1cos(x) + c2sin(x) + x cos(x) + sin(x)*ln|sec(x)|

#10. The characteristic equation for the LH is m2-9=0, which gives roots of m=3. The complementary solution to this DE is y=c1e3x+c2e-3x.
That tells us the particular solution is y = Ae3x + Be-3x.
The two equations we need to solve simultaneously are: A'e3x + B'e-3x = 0 and 3A'e3x + -3B'e-3x = 9xe-3x.
There are many ways to solve simultaneously. Whatever approach you choose, you should end up with:
A' = (3/2)xe-6x and B' = (-3/2)x.
Integrating gives us: A = (-1/24)e-6x-(1/4)xe-6x and B = (-3/4)x2
Thus our particular solution is: y = ((-1/24)e-6x-(1/4)xe-6x)*e3x + (-3/4)x2*e-3x , which simplifies to
y = (-1/24)e-3x- (1/4)xe-3x+ (-3/4)x2*e-3x.
Our final solution to the DE is:
y = c1e3x + c2e-3x + (-1/24)e-3x- (1/4)xe-3x+ (-3/4)x2*e-3x

 

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