How to solve DE with irreducible quadratic

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In summary, an irreducible quadratic in a differential equation is a term that cannot be factored into linear factors. To identify it, look for terms with a squared variable that cannot be written as a product of two linear terms. The general form of a solution to a DE with an irreducible quadratic is y = Ae^kx + Bxe^kx, where A and B are constants and k is a root of the irreducible quadratic term. To solve for the constants in the general solution, initial conditions must be used. The quadratic formula cannot be used to solve for x in a DE. Alternative methods for solving a DE with an irreducible quadratic include using the substitution x = e^t and the method of undetermined coefficients
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TOD
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I'm fine solving DE with a reducible quadratic, but what happens when the quadratic is irreducible, how would I go about solving the DE? So for example,
[tex]\frac{dP}{dt}=-P^{2}+10P-26[/tex]
Thanks in advance.
 
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  • #2
Try writing your DE as

[tex]
\frac{1}{-P^{2}+10P-26} \frac{dP}{dt}=1[/tex]


then complete the square for [itex]-P^{2}+10P-26[/itex], and try a trig substitution.
 

Related to How to solve DE with irreducible quadratic

1. How do I identify an irreducible quadratic in a differential equation?

An irreducible quadratic in a differential equation is a term that cannot be factored into linear factors. It will usually involve a term with a squared variable, such as x2 or y2. To identify it, look for terms with a squared variable that cannot be written as a product of two linear terms.

2. What is the general form of a solution to a DE with an irreducible quadratic?

The general form of a solution to a differential equation with an irreducible quadratic is y = Aekx + Bxekx, where A and B are constants and k is a root of the irreducible quadratic term. This form takes into account the two possible cases for the roots of an irreducible quadratic: real and distinct, and complex conjugates.

3. How do I solve for the constants in the general solution of a DE with an irreducible quadratic?

To solve for the constants A and B in the general solution, you will need to use initial conditions. These can be given in the problem or you can choose your own values. Substituting these values into the general solution will give you a system of equations that can be solved for the constants.

4. Can I use the quadratic formula to solve a DE with an irreducible quadratic?

No, the quadratic formula cannot be used to solve for x in a differential equation. The quadratic formula is used to solve for the roots of a quadratic equation, which is not the same as solving for a variable in a differential equation.

5. Are there any alternative methods for solving a DE with an irreducible quadratic?

Yes, there are other methods for solving a differential equation with an irreducible quadratic. One method is to use the substitution x = et, which can convert the DE into a linear equation. Another method is to use the method of undetermined coefficients, which involves guessing a particular solution and then solving for the constants. However, the general form of the solution will still be the same as mentioned in question 2.

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