anonymous
  • anonymous
FOR VARIATION OF PARAMETERS y"+y=csc^3xcotx
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
y2o2
  • y2o2
is it \[y'' + y = \csc^3 (x) \times \cot(x)\] ?
anonymous
  • anonymous
yes
y2o2
  • y2o2
so what is your question ?!\

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
can you solve it? the general solution?
anonymous
  • anonymous
can you solve it? :D
.Sam.
  • .Sam.
\[\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}\]
anonymous
  • anonymous
wolframalpha? i have the correct answer but i dont know how come it came with that answer..
anonymous
  • anonymous
the answer is y=C1cosx+C2sinx+1/6cotxcscx how come ?
.Sam.
  • .Sam.
? wolf's answer is same as yours
.Sam.
  • .Sam.
\[c_1 \cos (x)+c_2 \sin (x)+\frac{1}{6} \cos (x) \cot ^2(x)\]
.Sam.
  • .Sam.
you mean the cot^2(x?
anonymous
  • anonymous
i mean the cosxcot^2x
anonymous
  • anonymous
the correct answer is 1/6cotxcscx
.Sam.
  • .Sam.
Im not sure where's the mistake but we can ask someone to verify :) @Zarkon @TuringTest @Mertsj
anonymous
  • anonymous
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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.