y&nbsp;″+16y=−2sin⁡(4x)I try to solve this ode using the variation of parameters theorem.The characteristic polynomial of the homogenous equation is&nbsp;r2+16=0.Then&nbsp;u1(x)=sin⁡(4x),u2(x)=cos⁡(4x)y(x)=c1(x)sin⁡(4x)+c2(x)cos⁡(4x)I&nbsp;c1′(x)sin⁡(4x)+c2′cos⁡(4x)=0II&nbsp;c1′(x)cos⁡(4x)−c2′sin⁡(4x)=−2sin⁡(4x)Multiply I by&nbsp;cos⁡(4x)&nbsp;and II by sin(4x) and subtract.c1′(sin⁡(4x)cos⁡(4x)−cos⁡(4x)sin⁡(4x))+c2′(cos2(4x)+sin2(4x))=2We get&nbsp;c2′=2⇒c2=2x,c1'=−2cos⁡(4x)sin⁡(4x)=−2cot⁡(4x)⇒c1=−12ln|sin⁡(4x)|.I don't get why it incorrect, where am I wrong?

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y&amp;nbsp;″+16y=−2sin⁡(4x)I try to solve this ode using the variation of parameters theorem.The characteristic polynomial of the homogenous equation is&amp;nbsp;r2+16=0.Then&amp;nbsp;u1(x)=sin⁡(4x),u2(x)=cos⁡(4x)y(x)=c1(x)sin⁡(4x)+c2(x)cos⁡(4x)I&amp;nbsp;c1′(x)sin⁡(4x)+c2′cos⁡(4x)=0II&amp;nbsp;c1′(x)cos⁡(4x)−c2′sin⁡(4x)=−2sin⁡(4x)Multiply I by&amp;nbsp;cos⁡(4x)&amp;nbsp;and II by sin(4x) and subtract.c1′(sin⁡(4x)cos⁡(4x)−cos⁡(4x)sin⁡(4x))+c2′(cos2(4x)+sin2(4x))=2We get&amp;nbsp;c2′=2⇒c2=2x,c1&#039;=−2cos⁡(4x)sin⁡(4x)=−2cot⁡(4x)⇒c1=−12ln|sin⁡(4x)|.I don&#039;t get why it incorrect, where am I wrong?

Ali Price · Accepted Answer

Explanation:Your y(x) should be y(x)=c1(x)cos⁡(4x)+c2(x)sin⁡(4x).Using this, you will get two equations which are different from yours. So your second equation is wrong.Update: Firsty′(x)=c1′(x)cos⁡(4x)+c2′(x)sin⁡(4x)−4c1(x)sin⁡(4x)+4c2(x)cos⁡(4x).Let c1′(x)cos⁡(4x)+c2′(x)sin⁡(4x)=0.and then y′(x)=−4c1(x)sin⁡(4x)+4c2(x)cos⁡(4x).Now you can get y&#039;&#039;(x).

Chad Robinson · Accepted Answer

Step 1It seems like you are not paying enough attention to your differentiation. If you have u1,u2 as your homogeneous solution and look fory(x)=c1(x)u1(x)+c2(x)u2(x),then just taking one derivative gives:y′=(c1u1)′+(c2u2)′=c1′u1+c1u1′+c2′u2+c2u2′.Step 2The usual assumption (your equation I) makes is it so thaty′=c1u1′+c2u2′.The way you have written this right now, it reads likey′=d−2sin⁡4xwhich is your equation II and this is wrong. You take another derivative of y&#039; to find y&#039;&#039; and while you are doing so, you are going to use equation I again to obtain a simple expression. Then, you plug y and y&#039;&#039; to your original DE to obtain two sets of equations for c1 and c2.

\(\displaystyle{y}{''}+{16}{y}=-{\frac{{{2}}}{{{\sin{{\left({4}{x}\right)}}}}}}\) I try to solve this ode using

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