How do you solve a differential queation with derivatives inside a square on the denominator?C=f(x)y′1+y′2Where f(x) is some function in x and C is a constant.I am having a miserable time. I tried separation of variables but I end up getting a complex expression and that makes no sense:&nbsp;C1+y'2=f(x)y'&nbsp;⟺&nbsp;C2(1+y'2)=f2(x)y'2&nbsp;⟺&nbsp;B+By'2=f2(x)y'2&nbsp;⟺&nbsp;B+By'2−f2(x)y'2=0&nbsp;⟺&nbsp;y'2(B−f2(x))=−B&nbsp;⟺&nbsp;y'2=−BB−f2(x)&nbsp;⟺&nbsp;y'=−BB−f2(x)&nbsp;⟺&nbsp;dy=−BB−f2(x)dxWith&nbsp;B=C2.

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How do you solve a differential queation with derivatives inside a square on the denominator?C=f(x)y′1+y′2Where f(x) is some function in x and C is a constant.I am having a miserable time. I tried separation of variables but I end up getting a complex expression and that makes no sense:&amp;nbsp;C1+y&#039;2=f(x)y&#039;&amp;nbsp;⟺&amp;nbsp;C2(1+y&#039;2)=f2(x)y&#039;2&amp;nbsp;⟺&amp;nbsp;B+By&#039;2=f2(x)y&#039;2&amp;nbsp;⟺&amp;nbsp;B+By&#039;2−f2(x)y&#039;2=0&amp;nbsp;⟺&amp;nbsp;y&#039;2(B−f2(x))=−B&amp;nbsp;⟺&amp;nbsp;y&#039;2=−BB−f2(x)&amp;nbsp;⟺&amp;nbsp;y&#039;=−BB−f2(x)&amp;nbsp;⟺&amp;nbsp;dy=−BB−f2(x)dxWith&amp;nbsp;B=C2.

oanhtih6 · Accepted Answer

You can derive from the original equation that&amp;nbsp;|f(x)|&amp;gt;|C|&amp;nbsp;and that the sign of y&#039; is the product of the signs of f(x) and C. This allows you to determine the sign in the explicit equation for y&#039; pery&#039;(x)=sign(f(x))Cf(x)2-C2.This now is a quadrature problem, if that is solvable in symbolic expressions depends on the nature of f.

How do you solve a differential queation with

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