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anonymous

  • 5 years ago

for what values of the (real) constants a and b all solutions of the equations y"+ay'+by=0, are bounded for x between minus infinity and infinity?

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  1. helpmeplease
    • 5 years ago
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    Solve for the solution to the differential equation generally. Use the characteristic equation (e^(rx) substitution results). You should get the quadratic equation.

  2. anonymous
    • 5 years ago
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    helpme, i dont really get it. could u plis explain more in detail?

  3. helpmeplease
    • 5 years ago
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    That's a differential equation. You have to solve it. There are many different methods to determine the solutions. One of them is by substituting e^(rx) into the solution. You'll realize, that after you plug in the equation, that the e^(rx) divides into the zero. That leaves you with R^2+aR+b=0, which is a simple equation to solve. Just use the quadratic formula, with a and b as your coefficients. Later on, you substitute the solution back into your formula.

  4. anonymous
    • 5 years ago
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    helpmeplease is right, the Characteristic equation is\[r^2+ar+b=0\] so the roots are:\[r_{1}=\frac{-a+\sqrt{a^2-4b}}{2}, r_{2}=\frac{-a-\sqrt{a^2-4b}}{2}\]

  5. anonymous
    • 5 years ago
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    after get the roots, differentiate the solution then substitute them into y",y' and y in the original equation?

  6. anonymous
    • 5 years ago
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    if the roots are real distinct roots use: \[y=C_{1}e^{r_{1}x}+C_{1}e^{r_{2}x}\] If the roots are real and repeated where r1=r2 use: \[y=C_{1}xe^{r_{1}x}+C_{1}e^{r_{2}x}\] If the roots are Complex then use: \[\lambda=\frac{-a}{2}, \mu i=\pm \frac{\sqrt{a^2-4b}}{2}i\] \[y=e^{\lambda x}C_{1}\cos(\mu x)+e^{\lambda x}C_{2}\sin(\mu x)\]

  7. anonymous
    • 5 years ago
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    thanks nadeem and helpmeplease

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