I don"t think that"s right, however I"m not certain what else to do. And also where to go from there.
We have to use the initial values for the general solution$$y=c_1sin x+c_2cos x+y_p$$the details solution is$$y_p=Axsin x+Bxcos x$$$$A=0,B=frac12$$so the general solution is$$y=c_1sin x+c_2cos x+fracx2cos x$$now you can use the initial values to find the $A$ and also $B$
$y"" + y = -sin x$
solve the homogeneous equation
$y = c_1 sin t + c_2 cos x + y_p$
You are watching: Y'+y=sinx
$y = A xsin x + Bxcos x$
Why $A x sin x + B xcos x,$ and also not the more simple-minded $A sin x + B cos x$?
$A sin x + B cos x$ is already component of the homogeneous solution. So, when we plug it into the diff eq, that is going to "go away." We need something that when distinguished (twice) has actually a component that amounts to $-sin x$ yet is not in the homogeneous solution.
$y" = Asin x + Bcos x - B xsin x + Axcos x\y"" = -2B sin x + 2Acos x - A xsin x + Bxcos x\ y""+y = -2B sin x + 2Acos x = -sin x\B = frac 12, A = 0$
$y = c_1 sin t + c_2 cos x + frac 12 x cos x\y(0) = C_2 = 0\y"(0) = C_1 = 0$
$y = frac 12 xcos x$
edited Oct 18 "16 at 20:05
answered Oct 18 "16 in ~ 18:42
Doug MDoug M
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