exercise:B0046b0c67: Difference between revisions
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<math>\dydx = -ky</math>. The latter equation is similar to | <math>\dydx = -ky</math>. The latter equation is similar to | ||
<math>\dydx = y</math>, which has <math>e^x</math> for a solution. | <math>\dydx = y</math>, which has <math>e^x</math> for a solution. | ||
With this similarity in mind, | With this similarity in mind, it is not hard to guess, and then verify, that <math>y = e^{-kx}</math> is a solution to the original equation. | ||
it is not hard to guess, and then verify, | The problem is now to show that ''every'' solution is a constant multiple of | ||
that <math>y = e^{-kx}</math> is a solution to the original equation. | <math>e^{-kx}</math>. Prove this fact by assuming that <math>y = f(x)</math> is an arbitrary solution of <math>\dydx + ky = 0</math> and then showing that the derivative of the quotient <math>\frac{f(x)}{e^{-kx}}</math> is zero. | ||
The problem is now to show that ''every'' | |||
solution is a constant multiple of | |||
<math>e^{-kx}</math>. Prove this fact by assuming that <math>y = f(x)</math> | |||
is an arbitrary solution of | |||
<math>\dydx + ky = 0</math> and then showing that the derivative | |||
of the quotient <math>\frac{f(x)}{e^{-kx}}</math> is zero. | |||
Latest revision as of 23:02, 23 November 2024
[math]
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[/math]
An alternative approach to solving the linear differential equation [math]\dydx + ky = 0[/math] is to write it as [math]\dydx = -ky[/math]. The latter equation is similar to [math]\dydx = y[/math], which has [math]e^x[/math] for a solution. With this similarity in mind, it is not hard to guess, and then verify, that [math]y = e^{-kx}[/math] is a solution to the original equation. The problem is now to show that every solution is a constant multiple of [math]e^{-kx}[/math]. Prove this fact by assuming that [math]y = f(x)[/math] is an arbitrary solution of [math]\dydx + ky = 0[/math] and then showing that the derivative of the quotient [math]\frac{f(x)}{e^{-kx}}[/math] is zero.