78_pdfsam_math 54 differential equation solutions odd

78_pdfsam_math 54 differential equation solutions odd - | x...

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Chapter 2 Isolating dy/dx ,weobta in dy dx = y 2 y 2 x x = y x 2 y 2 . The right-hand side cannot be factorized as p ( x ) q ( y ), and so the equation is not separable. Also, it is not linear with y as the dependent variable (because of 2 y 2 term). By taking the reciprocals we also conclude that it is not linear with the dependent variable x . 7. The equation (3 x 2 + y ) dx +( x 2 y x ) dy = 0 is not separable or linear. To see if it is exact, we compute ∂M ∂y =1 6 =2 xy 1= ∂N ∂x . Thus, the equation is not exact. To see if we can Fnd an integrating factor, we compute ∂M/∂y ∂N/∂x N = 2 2 xy x 2 y x = 2( xy 1) x ( xy 1) = 2 x . ±rom this we see that the integrating factor will be µ ( x )=exp ±Z 2 x dx ² =exp(
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Unformatted text preview: | x | ) = x − 2 . To solve the equation, we multiply it by the integrating factor x − 2 to obtain (3 + yx − 2 ) dx + ( y − x − 1 ) dy = 0 . This is now exact. Thus, we want to Fnd F ( x, y ). To do this, we integrate M ( x, y ) = 3+ yx − 2 with respect to x to get F ( x, y ) = Z ( 3 + yx − 2 ) dx = 3 x − yx − 1 + g ( y ) ⇒ F y ( x, y ) = − x − 1 + g ( y ) = N ( x, y ) = y − x − 1 ⇒ g ( y ) = y ⇒ g ( y ) = y 2 2 . Therefore, F ( x, y ) = 3 x − yx − 1 + y 2 2 . And so we see that an implicit solution is y 2 2 − y x + 3 x = C. 74...
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