- #1
misogynisticfeminist
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There is an example in a book regarding DEs which I do not understand. Solve the IVP
[tex] y'=y^2-4, y(0)=-2 [/tex]where t is the independent variable
We first solve by separation of variables to arrive at the 1-parameter solution.
[tex] -\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c[/tex]
Simplifying and expressing the solution explicitly, we find that,
[tex] y=2\frac{1+ce^{4t}}{1-ce^{4t}} [/tex]
Taking the initial condition,
[tex] -2=2\frac{1+c}{1-c} [/tex] which simplifies to,
-1=1.
They said that the solution is wrong because:
we can express the DE as, [tex] y'=(y+2)(y-2) [/tex] and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?
[tex] y'=y^2-4, y(0)=-2 [/tex]where t is the independent variable
We first solve by separation of variables to arrive at the 1-parameter solution.
[tex] -\frac{1}{4}ln (y+2)+\frac{1}{4}ln (y-2)=t+c[/tex]
Simplifying and expressing the solution explicitly, we find that,
[tex] y=2\frac{1+ce^{4t}}{1-ce^{4t}} [/tex]
Taking the initial condition,
[tex] -2=2\frac{1+c}{1-c} [/tex] which simplifies to,
-1=1.
They said that the solution is wrong because:
we can express the DE as, [tex] y'=(y+2)(y-2) [/tex] and that the when y=-2, and y=2 satisfies this equation (what does it mean?). How do we "preclude" y=-2 and y=-2 before solving starting to solve the DE?
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