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Unformatted text preview: Exercises 2.4
15. This diﬀerential equation is expressed in the variables r and θ. Since the variables x and y are dummy variables, this equation is solved in exactly the same way as an equation in x and y . We will look for a solution with independent variable θ and dependent variable r . We see that the diﬀerential equation is expressed in the diﬀerential form M (r, θ) dr + N (r, θ) dθ = 0, This implies that Mθ (r, θ) = − sin θ = Nr (r, θ), and so the equation is exact. Therefore, to solve the equation we need to ﬁnd a function F (r, θ) that has cos θ dr + (−r sin θ + eθ ) dθ as its total diﬀerential. Integrating M (r, θ) with respect to r we see that F (r, θ) = ⇒ Thus we have that g (θ) = eθ ⇒ g (θ) = eθ , cos θ dr = r cos θ + g (θ) where M (r, θ) = cos θ and N (r, θ) = −r sin θ + eθ . Fθ (r, θ) = −r sin θ + g (θ) = N (r, θ) = −r sin θ + eθ . where the constant of integration will be incorporated into the parameter of the solution. Substituting this expression for g (θ) into the expression we found for F (r, θ) yields F (r, θ) = r cos θ + eθ . From this we see that the solution is given by the one parameter family r cos θ + eθ = C , or, solving for r , r= C − eθ = (C − eθ ) sec θ. cos θ 17. Partial derivatives of M (x, y ) = 1/y and N (x, y ) = − (3y − x/y 2 ) are ∂ ∂M = ∂y ∂y 1 y =− 1 y2 and ∂ ∂N = ∂x ∂x −3y + x y2 = 1 . y2 Since ∂M/∂y = ∂N/∂x, the equation is not exact. 63 ...
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