Absolute signs in solving ODEs1ydydx=1xThe way I solve this:for all x in (−∞,0)ln|y|=ln|x|+A where A is a real constant (since we know antiderivatives are separated by a constant)|y|=B|x| where B is a positive constant equals exp(A)y=Bx or −BxIt seems to me y=Cx where C is a real non-zero constant that equals either B or -B.Since it seems like if y is equal to + or −Bx for the entire interval (−∞,0), then it must be only +Bx or −Bx i.e. Cx for some subinterval. If it is +Bx, then for values outside and near this interval, they must also be +Bx since we know y is continuous and x is not zero. Vice versa.But I don't know how to prove this. How to prove y=Cx for (−∞,0)?If that is the case, then I can write the general form of the differential equation picewisely as y=C1 for (−∞,0),C2x for (0,∞), where C1 and C2 are non-zero reals.

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Absolute signs in solving ODEs1ydydx=1xThe way I solve this:for all x in (−∞,0)ln|y|=ln|x|+A where A is a real constant (since we know antiderivatives are separated by a constant)|y|=B|x| where B is a positive constant equals exp(A)y=Bx or −BxIt seems to me y=Cx where C is a real non-zero constant that equals either B or -B.Since it seems like if y is equal to + or −Bx for the entire interval (−∞,0), then it must be only +Bx or −Bx i.e. Cx for some subinterval. If it is +Bx, then for values outside and near this interval, they must also be +Bx since we know y is continuous and x is not zero. Vice versa.But I don&#039;t know how to prove this. How to prove y=Cx for (−∞,0)?If that is the case, then I can write the general form of the differential equation picewisely as y=C1 for (−∞,0),C2x for (0,∞), where C1 and C2 are non-zero reals.

Brock Floyd · Accepted Answer

Step 1The first point of confusion here is that you antidifferentiatied&amp;nbsp;y′y=1xto&amp;nbsp;ln⁡(|y|)=ln⁡(|x|)+A,but this is not entirely correct. The antiderivatives of&amp;nbsp;1t&amp;nbsp;are given by the piecewise,ln⁡(−t)+A;,∀{t&amp;lt;0}ln⁡(t)+A;,∀{t&amp;gt;0}.Taking this into account, you should haveln⁡(−y)=ln⁡(−x)+Aor&amp;nbsp;ln⁡(y)=ln⁡(−x)+B.This translates to&amp;nbsp;y(x)=exp⁡(A)xor&amp;nbsp;y(x)=−exp⁡(B)x.Step 2The second point of is in interpreting the &quot;or.&quot; The two solutions we have here are that&amp;nbsp;y(x)=c0x&amp;nbsp;with&amp;nbsp;c0&amp;gt;0, or that&amp;nbsp;y(x)=c1x&amp;nbsp;with&amp;nbsp;c1&amp;lt;0. You could not have both happening simultaneously in all of&amp;nbsp;(−∞,0). Your question is, how do you prove it? Well, notice that the equation must be satisfied everywhere. But it cannot be satisfied if there are jump discontinities, which occur if you have a constant of one kind in one subinterval, and a constant of some other kind everywhere else. You can show that if    f    (    x    )    =          {                                    A            x                                x            ≤                          C                                                            B            x                                x            &amp;gt;                          C                                                  ,  then f is not even differentiable at C unless&amp;nbsp;A=B. You can prove it just using the definition of the derivative. The method for solving the equation, when done properly, already implies this, though, so in practice, you would not need to prove it.

Absolute signs in solving ODEs \(\displaystyle{\frac{{{1}}}{{{y}}}}{\frac{{\text{d}{y}}}{{\text{d}{x}}}}={\frac{{{1}}}{{{x}}}}\) The way I

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