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9_pdfsam_math 54 differential equation solutions odd

# 9_pdfsam_math 54 differential equation solutions odd - and...

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Exercises 1.2 So, dx dt + tx = 2 sin 2 t + t cos 2 t sin 2 t on any interval. Therefore, x ( t ) is not a solution to the given differential equation. 7. We differentiate y = e 2 x 3 e x twice: dy dx = d dx ( e 2 x 3 e x ) = e 2 x (2) 3 e x ( 1) = 2 e 2 x + 3 e x ; d 2 y dx 2 = d dx dy dx = d dx ( 2 e 2 x + 3 e x ) = 2 e 2 x (2) + 3 e x ( 1) = 4 e 2 x 3 e x . Substituting y , y , and y into the differential equation and collecting similar terms, we get d 2 y dx 2 dy dx 2 y = ( 4 e 2 x 3 e x ) ( 2 e 2 x + 3 e x ) 2 ( e 2 x 3 e x ) = (4 2 2) e 2 x + ( 3 3 2( 3)) e x = 0 . Hence y = e 2 x 3 e x is an explicit solution to the given differential equation. 9. Differentiating the equation x 2 + y 2 = 6 implicitly, we obtain 2 x + 2 yy = 0 y = x y . Since there can be no function y = f ( x ) that satisfies the differential equation
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Unformatted text preview: and the diFerential equation y = − x/y on the same interval , we see that x 2 + y 2 = 6 does not de±ne an implicit solution to the diFerential equation. 11. DiFerentiating the equation e xy + y = x − 1 implicitly with respect to x yields d dx ( e xy + y ) = d dx ( x − 1) ⇒ e xy d dx ( xy ) + dy dx = 1 ⇒ e xy ± y + x dy dx ² + dy dx = 1 ⇒ ye xy + dy dx ( xe xy + 1) = 1 ⇒ dy dx = 1 − ye xy 1 + xe xy = e xy ( e − xy − y ) e xy ( e − xy + x ) = e − xy − y e − xy + x . 5...
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