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64_pdfsam_math 54 differential equation solutions odd

# 64_pdfsam_math 54 differential equation solutions odd - r...

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Chapter 2 and conclude that it is not separable because the right-hand side cannot be represented as a product of two functions of single variables x and y . Also, the right-hand side is not linear with respect to y which implies that the equation is not linear with y as the dependent variable. Similarly, choosing x as the dependent variable (taking the reciprocals of both sides) we conclude that the equation is not linear either. 5. The differential equation is not separable because (2 xy + cos y ) cannot be factored. This equation can be put in standard form by defining x as the dependent variable and y as the independent variable. This gives dx dy + 2 y x = cos y y 2 , so we see that the differential equation is linear. If we set M ( x, y ) = y 2 and N ( x, y ) = 2 xy + cos y we are able to see that the differential equation is also exact because M y ( x, y ) = 2 y = N x ( x, y ). 7. In this problem, the variables are
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Unformatted text preview: r and θ , M ( r, θ ) = θ , and N ( r, θ ) = 3 r − θ − 1. Because ∂M ∂θ = 1 6 = 3 = ∂N ∂r , the equation is not exact. With r as the dependent variable, the equation takes the form dr dθ = − 3 r − θ − 1 θ = − 3 θ r + θ + 1 θ , and it is linear. Since the right-hand side in the above equation cannot be factored as p ( θ ) q ( r ), the equation is not separable. 9. We have that M ( x, y ) = 2 xy + 3 and N ( x, y ) = x 2 − 1. Therefore, M y ( x, y ) = 2 x = N x ( x, y ) and so the equation is exact. We will solve this equation by ±rst integrating M ( x, y ) with respect to x , although integration of N ( x, y ) with respect to y is equally easy. Thus F ( x, y ) = Z (2 xy + 3) dx = x 2 y + 3 x + g ( y ) . 60...
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