#4. The complementary solution to this DE is y=c_{1}cos(x) + c_{2}sin(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' = -tan^{2}x = 1 - sec^{2}x 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 = c_{1}cos(x) + c_{2}sin(x) + x cos(x) + sin(x)*ln|sec(x)|

#10. The characteristic equation for the LH is m^{2}-9=0, which gives roots of m=±3. The complementary solution to this DE is y=c_{1}e^{3x}+c_{2}e^{-3x}.

That tells us the particular solution is y = Ae^{3x} + Be^{-3x}.

The two equations we need to solve simultaneously are: A'e^{3x} + B'e^{-3x} = 0 and 3A'e^{3x} + -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)x^{2}

Thus our particular solution is: y = ((-1/24)e^{-6x}-(1/4)xe^{-6x})^{}*e^{3x} + (-3/4)x^{2}*e^{-3x} , which simplifies to

y = (-1/24)e^{-3x}- (1/4)xe^{-3x}+ (-3/4)x^{2}*e^{-3x}.

Our final solution to the DE is:

y = c_{1}e^{3x} + c_{2}e^{-3x} + (-1/24)e^{-3x}- (1/4)xe^{-3x}+ (-3/4)x^{2}*e^{-3x}