Separable First Order Differential Equation

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tinopham
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Homework Statement



[itex]\frac{dy}{dx} = y \sqrt{x}[/itex], f(9) = 5

The Attempt at a Solution



[itex]\int dy/y = \int \sqrt{x} dx[/itex]

[itex]ln |y| = \frac{2}{3} x^\frac{3}{2} + c[/itex]

[itex]y = e^{\frac{2}{3}x^\frac{3}{2}} + C[/itex]

[itex]y = Ce^{\frac{2}{3}x^\frac{3}{2}}[/itex]

[itex]5 = Ce^{\frac{2}{3}9^\frac{3}{2}}[/itex]

[itex]5 = Ce^{18}[/itex]

[itex]C = \frac{5}{e^{18}}[/itex]

Thus,[itex]y = \frac{5}{e^{18}} e^{\frac{2}{3}x^\frac{3}{2}}[/itex]
[itex]y = 5e^{-18} e^{\frac{2}{3}x^\frac{3}{2}}[/itex]
[itex]y = 5 e^{\frac{2}{3}x^\frac{3}{2}-18}[/itex]
 
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tinopham said:

Homework Statement



[itex]\frac{dy}{dx} = y \sqrt{x}[/itex], f(9) = 5

The Attempt at a Solution



[itex]\int dy/y = \int \sqrt{x} dx[/itex]

[itex]ln |y| = \frac{2}{3} x^\frac{3}{2} + c[/itex]

[itex]y = e^{\frac{2}{3}x^\frac{3}{2}} + C[/itex]

[itex]y = Ce^{\frac{2}{3}x^\frac{3}{2}}[/itex]

[itex]5 = Ce^{\frac{2}{3}9^\frac{3}{2}}[/itex]

[itex]5 = Ce^{18}[/itex]

[itex]C = \frac{5}{e^{18}}[/itex]

Thus,[itex]y = \frac{5}{e^{18}} e^{\frac{2}{3}x^\frac{3}{2}}[/itex]
[itex]y = 5e^{-18} e^{\frac{2}{3}x^\frac{3}{2}}[/itex]
[itex]y = 5 e^{\frac{2}{3}x^\frac{3}{2}-18}[/itex]
Hello tinopham. Welcome to PF !

Do you have a question about this ?
 
Hi SamS, I was going to ask a question, but I was able to solve it. Thanks!
 
kushan said:
have ln y + ln c
1) The integral of 1/y is ln|y|+ c, not ln y.
2) Since tinopham had a "c" on the right side ogf the equation, it is not necessary to have a constant on the left. The two constants of integration can be combined on oneside.