Question

FOR VARIATION OF PARAMETERS

y"+y=csc^3xcotx

Answer 1

is it
\[y'' + y = \csc^3 (x) \times \cot(x)\]
?

Answer 2

yes

Answer 3

so what is your question ?!\

Answer 4

can you solve it? the general solution?

Answer 5

can you solve it?
:D

Answer 6

\[\begin{array}{l} \text{Solve }\frac{d^2y(x)}{dx^2}+y(x)=\cot (x) \csc ^3(x): \\ \text{The general solution will be the \sum of the complementary solution and particular solution.} \\ \text{Find the complementary solution by solving }\frac{d^2y(x)}{dx^2}+y(x)\text{ = }0: \\ \text{Assume a solution }\text{}\text{will be proportional \to }e^{\lambda x}\text{ for some constant }\lambda . \\ \text{Substitute }y(x)\text{ = }e^{\lambda x}\text{ into the differential equation:} \\ \frac{d^2\text{}}{dx^2}\left(e^{\lambda x}\right)+e^{\lambda x}\text{ = }0 \\ \text{Substitute }\frac{d^2\text{}}{dx^2}\left(e^{\lambda x}\right)\text{ = }\lambda ^2 e^{\lambda x}: \\ \lambda ^2 e^{\lambda x}+e^{\lambda x}\text{ = }0 \\ \text{Factor out }e^{\lambda x}: \\ \left(\lambda ^2+1\right) e^{\lambda x}\text{ = }0 \\ \text{Since }e^{\lambda x}\neq 0\text{ for any finite }\lambda \text{, the zeros must come from the polynomial:} \\ \lambda ^2+1\text{ = }0 \\ \text{Solve for }\lambda : \\ \lambda =i\text{ or }\lambda =-i \\ \text{The roots }\lambda \text{ = }\text{$\pm $ }i\text{ give }y_1(x)=c_1 e^{i x}\text{, }y_2(x)=c_2 e^{-i x}\text{ as solutions, where }c_1\text{ and }c_2\text{ are arbitrary constants.} \\ \text{The general solution is the \sum of the \above solutions:} \\ y(x)\text{ = }y_1(x)+y_2(x)\text{ = }c_1 e^{i x}+c_2 e^{-i x} \\ \text{Apply Euler's identity }e^{\alpha +i \beta }=e^{\alpha } \cos (\beta )+i e^{\alpha } \sin (\beta ): \\ y(x)\text{ = }c_1 (\cos (x)+i \sin (x))+c_2 (\cos (x)-i \sin (x)) \\ \text{Regroup terms:} \\ y(x)\text{ = }\left(c_1+c_2\right) \cos (x)+i \left(c_1-c_2\right) \sin (x) \\ \text{Redefine }c_1+c_2\text{ as }c_1\text{ and }i \left(c_1-c_2\right)\text{ as }c_2\text{, since these are arbitrary constants:} \\ y(x)\text{ = }c_1 \cos (x)+c_2 \sin (x) \\ \text{Determine the particular solution \to }\frac{d^2y(x)}{dx^2}+y(x)\text{ = }\cot (x) \csc ^3(x)\text{ by variation of parameters:} \\ \text{List the basis solutions \in }y_{\text{c}}(x): \\ y_{b_1}(x)=\cos (x)\text{ and }y_{b_2}(x)=\sin (x) \\ \text{Compute the Wronskian of }y_{b_1}(x)\text{ and }y_{b_2}(x): \\ \mathcal{W}(x)\text{ = }\left|\begin{array}{cc} \cos (x) & \sin (x) \\ \frac{d\cos (x)}{dx} & \frac{d\sin (x)}{dx} \\\end{array}\right|\text{ = }\left|\begin{array}{cc} \cos (x) & \sin (x) \\ -\sin (x) & \cos (x) \\\end{array}\right|\text{ = }1 \\ \text{Let }f(x)=\cot (x) \csc ^3(x): \\ \text{Let }v_1(x)=-\int\limits \frac{f(x) y_{b_2}(x)}{\mathcal{W}(x)} \, dx\text{ and }v_2(x)=\int\limits \frac{f(x) y_{b_1}(x)}{\mathcal{W}(x)} \, dx: \\ \text{The particular solution will be given by:} \\ y_p(x)\text{ = }v_1(x) y_{b_1}(x)+v_2(x) y_{b_2}(x) \\ \text{Compute }v_1(x): \\ v_1(x)\text{ = }-\int\limits \cot (x) \csc ^2(x) \, dx\text{ = }\frac{\cot ^2(x)}{2} \\ \text{Compute }v_2(x): \\ v_2(x)\text{ = }\int\limits \cot ^2(x) \csc ^2(x) \, dx\text{ = }-\frac{1}{3} \cot ^3(x) \\ \text{The particular solution is thus:} \\ y_p(x)\text{ = }v_1(x) y_{b_1}(x)+v_2(x) y_{b_2}(x)\text{ = }\frac{1}{2} \cos (x) \cot ^2(x)-\frac{1}{3} \cos (x) \cot ^2(x) \\ \text{Simplify:} \\ y_p(x)\text{ = }\frac{1}{6} \cos (x) \cot ^2(x) \\ \text{The general solution is given by:} \\ y(x)\text{ = }y_{\text{c}}(x)\text{ + }y_p(x)\text{ = }c_1 \cos (x)+c_2 \sin (x)+\frac{1}{6} \cos (x) \cot ^2(x) \\\end{array}\]

Answer 7

wolframalpha? i have the correct answer but i dont know how come it came with that answer..

Answer 8

the answer is y=C1cosx+C2sinx+1/6cotxcscx how come ?

Answer 9

? wolf's answer is same as yours

Answer 10

\[c_1 \cos (x)+c_2 \sin (x)+\frac{1}{6} \cos (x) \cot ^2(x)\]

Answer 11

you mean the cot^2(x?

Answer 12

i mean the cosxcot^2x

Answer 13

the correct answer is 1/6cotxcscx

Answer 14

Im not sure where's the mistake but we can ask someone to verify :)
@Zarkon @TuringTest @Mertsj

Answer 15

turing test give the idea about the wolframalpha i hope they can help me now. :))) how about the reduction of order? the wolframalpha's answer is wrong. tsk.