- #1

- 363

- 0

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?

Last edited: