Find p(t) for the ODE given that the Wroian is a non-zero constant.Consider an ODE of the form&nbsp;d2x&nbsp;dt&nbsp;2+p(t)&nbsp;dx&nbsp;&nbsp;dt&nbsp;+q(t)x=0Suppose that we have two solutions&nbsp;x1(t)&nbsp;and&nbsp;x2(t)&nbsp;to this ODE and their Wroian is a non-zero constant. That is,W[x1,x2]=C≠0What is p(t)? (Find an explicit formula).The formula for the Wroian isW[x1,x2]=x1&nbsp;dx&nbsp;2&nbsp;dt&nbsp;−x2&nbsp;dx&nbsp;1&nbsp;dt&nbsp;Since the Wroian is non-zero, we know that x1 and x2 are independent. This means that&nbsp;&nbsp;dx&nbsp;1&nbsp;dt&nbsp;&nbsp;and&nbsp;&nbsp;dx&nbsp;2&nbsp;dt&nbsp;&nbsp;are such that all variables cancel out. I'm not sure how to proceed from here. Can I get some hints on how to find p(t)? ThanksEdit: Using Abel's theorem, we know that if&nbsp;x1&nbsp;and&nbsp;x2&nbsp;are solutions to the ODE thenW[x1,x2]=Ae−∫p(t)&nbsp;dt&nbsp;Then this can be used to solve for p(t).W[x1,x2]&nbsp;=Ae−∫p(t)dt=C&nbsp;e−∫p(t)dt=CA&nbsp;&nbsp;∫p(t)dt=−ln⁡(CA)&nbsp;p(t)=ddt(−ln⁡(CA))&nbsp;&nbsp;=0So&nbsp;p(t)=0.

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Find p(t) for the ODE given that the Wroian is a non-zero constant.Consider an ODE of the form&amp;nbsp;d2x&amp;nbsp;dt&amp;nbsp;2+p(t)&amp;nbsp;dx&amp;nbsp;&amp;nbsp;dt&amp;nbsp;+q(t)x=0Suppose that we have two solutions&amp;nbsp;x1(t)&amp;nbsp;and&amp;nbsp;x2(t)&amp;nbsp;to this ODE and their Wroian is a non-zero constant. That is,W[x1,x2]=C≠0What is p(t)? (Find an explicit formula).The formula for the Wroian isW[x1,x2]=x1&amp;nbsp;dx&amp;nbsp;2&amp;nbsp;dt&amp;nbsp;−x2&amp;nbsp;dx&amp;nbsp;1&amp;nbsp;dt&amp;nbsp;Since the Wroian is non-zero, we know that x1 and x2 are independent. This means that&amp;nbsp;&amp;nbsp;dx&amp;nbsp;1&amp;nbsp;dt&amp;nbsp;&amp;nbsp;and&amp;nbsp;&amp;nbsp;dx&amp;nbsp;2&amp;nbsp;dt&amp;nbsp;&amp;nbsp;are such that all variables cancel out. I&#039;m not sure how to proceed from here. Can I get some hints on how to find p(t)? ThanksEdit: Using Abel&#039;s theorem, we know that if&amp;nbsp;x1&amp;nbsp;and&amp;nbsp;x2&amp;nbsp;are solutions to the ODE thenW[x1,x2]=Ae−∫p(t)&amp;nbsp;dt&amp;nbsp;Then this can be used to solve for p(t).W[x1,x2]&amp;nbsp;=Ae−∫p(t)dt=C&amp;nbsp;e−∫p(t)dt=CA&amp;nbsp;&amp;nbsp;∫p(t)dt=−ln⁡(CA)&amp;nbsp;p(t)=ddt(−ln⁡(CA))&amp;nbsp;&amp;nbsp;=0So&amp;nbsp;p(t)=0.

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Explanation:If&amp;nbsp;xi&amp;nbsp;″+pxi′+qxi=0,u=1,2, thenx1(x2&#039;&#039;+px2&#039;+qx2)=x2(x1&#039;&#039;+px1&#039;+qx1)=0→&amp;nbsp;(x1x2&#039;&#039;−x1&#039;&#039;x2)+p(x1x2&#039;−x2x1&#039;)=0→&amp;nbsp;p=−x1x2&#039;&#039;−x1&#039;&#039;x2x1x2&#039;−x2x1&#039;=−x1x2&#039;&#039;−x1&#039;&#039;x2W[x1,x2]

Find p(t) for the ODE given that the

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