Differential equations: Shouldn't the method of separation of variables miss the solutions when x=0?Consider the ODE:xdydx=2y (1)We apply the separation of variables technique to obtain:1ydy=2xdx (2)which provides the general solution:y(x)=Cx2The book mentions that "the general solution (3) satisfies (1) for any value of the constant C and for all values of the variable x".However, I believe the statement is wrong when x=0. By using the separation of variables, we divided by x and y to obtain (2). Thus, we make the implicit assumption that x≠0, and the separation of variables will generate solutions where x≠0.How can y be defined for any x, when we implicitly assume that x≠0?

Question

Differential equations: Shouldn&#039;t the method of separation of variables miss the solutions when x=0?Consider the ODE:xdydx=2y (1)We apply the separation of variables technique to obtain:1ydy=2xdx (2)which provides the general solution:y(x)=Cx2The book mentions that &quot;the general solution (3) satisfies (1) for any value of the constant C and for all values of the variable x&quot;.However, I believe the statement is wrong when x=0. By using the separation of variables, we divided by x and y to obtain (2). Thus, we make the implicit assumption that x≠0, and the separation of variables will generate solutions where x≠0.How can y be defined for any x, when we implicitly assume that x≠0?

slokningol09 · Accepted Answer

You are correct, as an ordinary DE the domain does not contain the vertical line x=0. It is a special feature of this equation that you can take two solutions =C1x2 for x&amp;lt;0 and =C2x2 for x&amp;gt;0 and connect them to a differentiable function at x=0. But there is no innate connection between the two constants, one can take any random combination.

Dixie Reed · Accepted Answer

Step 1For a differential equation of the form&amp;nbsp;α(x)y′(x)=f(x)(g∘{y})(x)&amp;nbsp;in y, y need not be differentiable at any points such that&amp;nbsp;α(x)=0&amp;nbsp;in order for the equation to be satisfied everywhere else. Hence, the domain of y must initially be assumed to be&amp;nbsp;ℝ∖Zα, where&amp;nbsp;Zα&amp;nbsp;={x∈R:α(x)=0}, and it is only extended to&amp;nbsp;R&amp;nbsp;if the singularities are removable.In your example equation,&amp;nbsp;xy′(x)=2y(x), y is assumed implicitly to have domain&amp;nbsp;R∖{0}, making the domain disconnected. The equation must be separately solved for the cases&amp;nbsp;x&amp;lt;0&amp;nbsp;and&amp;nbsp;x&amp;gt;0. For the case of&amp;nbsp;x&amp;lt;0, there are three further cases to consider:&amp;nbsp;y&amp;lt;0,y=0, which is trivial, and&amp;nbsp;y&amp;gt;0. For the three cases, one hasln⁡[−y(x)]=2ln⁡(−x)+C−&amp;nbsp;y(x)=0ln⁡[y(x)]=2ln⁡(−x)+C−&amp;nbsp;,respectively. These simplify to&amp;nbsp;y(x)=−C−x2y(x)=0x2y(x)=C−x2,which are the cases such thaty(x)=A−x2,,A−&amp;lt;0y(x)=A−x2,,A≡0y(x)=A−x2,,A−&amp;gt;0.Step 2The cases can be collapsed into a single case, where&amp;nbsp;A−∈R, and&amp;nbsp;y(x)=A−x2.For&amp;nbsp;x&amp;gt;0, the analysis is completely analogous, resulting in the most general solution to be&amp;nbsp;y(x)=A+x2&amp;nbsp;where&amp;nbsp;A+∈R. Thus, the general solution family for the equation is given byy(x)={A−x2x&amp;lt;0A+x2x&amp;gt;0Of course, some solutions can be continuously and differentiably extended to&amp;nbsp;R, but this requires that&amp;nbsp;A≡A+=A. Such solutions satisfy the differential equation in&amp;nbsp;R.

Differential equations: Shouldn't the method of separation of

Answered question

Answer & Explanation

New Questions in Calculus 2