y"+2y'+y=sinx
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Your work is very neat - but very difficult to follow due to the orientation of snap-shot.
Can you please try to re-post with correct orientation?
Can you please try to re-post with correct orientation?
HallsofIvy
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The differential equation is y''+ 2y'+ y= A sin(x). The associated homogeneous equation is y''+ 2y'+ y= 0 which has characteristic equation . That has the single solution r= -1 so the general solution to the homogeneous equation is .
You appear to have done that part correctly. Now, knowing that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x), I would look for a solution to the entire equation for constants A and B.
y'(x)= A cos(x)- B sin(x) and y''(x)= -A sin(x)- B cos(x) so that y''+ 2y'+ y= -A sin(x)- B cos(x)+ 2A cos(x)- 2B sin(x)+ A sin(x)+ B cos(x)= 2B sin(x)+ 2A cos(x)= sin(x). Since this is to be true for all x, we must have A= 0 and B= 1/2.`
Were you trying to use the "variation of parameters" method? That works but, here, is extremely tedious and error-prone. In this case the "right hand side" is sin(x) which is one of the kinds of functions we expect to get as a solution to a "linear equation with constant coefficients". When the "right hand side" is an exponential, a polynomial, sine or cosine, or combinations of those, try "undetermined coefficients" rather than "variation of parameters".
You appear to have done that part correctly. Now, knowing that the derivative of sin(x) is cos(x) and that the derivative of cos(x) is -sin(x), I would look for a solution to the entire equation for constants A and B.
y'(x)= A cos(x)- B sin(x) and y''(x)= -A sin(x)- B cos(x) so that y''+ 2y'+ y= -A sin(x)- B cos(x)+ 2A cos(x)- 2B sin(x)+ A sin(x)+ B cos(x)= 2B sin(x)+ 2A cos(x)= sin(x). Since this is to be true for all x, we must have A= 0 and B= 1/2.`
Were you trying to use the "variation of parameters" method? That works but, here, is extremely tedious and error-prone. In this case the "right hand side" is sin(x) which is one of the kinds of functions we expect to get as a solution to a "linear equation with constant coefficients". When the "right hand side" is an exponential, a polynomial, sine or cosine, or combinations of those, try "undetermined coefficients" rather than "variation of parameters".
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The image is sideways, at least on my end, and difficult to read otherwise. The first thing I notice is we have the characteristic root of multiplcity 2, thus the homogeneous solution is:
[MATH]y_h(x)=c_1e^{-x}+c_2xe^{-x}=\left(c_1+c_2x\right)e^{-x}[/MATH]
Are you with me so far?
[MATH]y_h(x)=c_1e^{-x}+c_2xe^{-x}=\left(c_1+c_2x\right)e^{-x}[/MATH]
Are you with me so far?
