- #1
Jeff12341234
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I'm not sure if my answer is correct. Did I make a mistake somewhere? I'm not sure the ± needs to be there.
Correcting Mistakes in Representing Constants for a Differential Equation?
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SUMMARY
The discussion centers on correcting the representation of constants in a differential equation. The user initially questioned the necessity of the ± sign and the derivation of a quadratic equation from the given equations. The equations presented are y(2) = (C_1 + 6C_2)e^6 = e^6 and y'(1) = (3C_1 + 4C_2)e^3 = e^3, leading to two linear equations: C_1 + 6C_2 = 1 and 3C_1 + 4C_2 = 1. The user acknowledged an error in omitting the + sign in the derivative, ultimately determining that C_1 = -1 and C_2 = 1.
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- #2
HallsofIvy
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How in the world did you get a quadratic equation out of this? [itex]y(2)= (C_1+ 6C_2)e^6= e^6[/itex] and [itex]y'(1)= (3C_1+ 4C_2)e^3= e^3[/itex]. The derivative is [itex]y'= 3C_1e^{3x}+ C_2e^{3x}+ 3C_2xe^{3x}= ([3C_1+ C_2]+ 3C_2x)e^x[/itex]. It does not involve "[itex]C_1C_2[/itex]"!
You have [itex]C_1+ 6C_2= 1[/itex] and [itex]3C_1+ 4C_2= 1[/itex], two linear equations.
You have [itex]C_1+ 6C_2= 1[/itex] and [itex]3C_1+ 4C_2= 1[/itex], two linear equations.
- #3
Jeff12341234
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c1 is represented by c, c2 is represented by d
That's y'
I did make an error by leaving out the + sign between c1 and c2 for y'
That makes c1 = -1 and c2 = 1
That's y'
I did make an error by leaving out the + sign between c1 and c2 for y'
That makes c1 = -1 and c2 = 1
Last edited:
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