2302-practice-mid1-soln

# C y dy dx arctan x y y 1 1 solution to use the

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c. y dy dx = arctan( x + y ); y (1) = 1. Solution: To use the Existence and Uniqueness Theorem for first-order differential equations, we must first put this equation in the form dy dx = f ( x, y ) . In this case we get dy dx = arctan( x + y ) y . Now we test whether f and ∂f/∂y are continuous in some neighborhood of ( x 0 , y 0 ) = (1 , 1). The only discontinuity of f is along the line y = 0, so f is certainly continuous in a neighborhood of (1 , 1). Using the quotient rule, we see that ∂f/∂y is ∂f ∂y = y ( x + y ) 2 +1 arctan( x + y ) y 2 = 1 y (( x + y ) 2 + 1 arctan( x + y ) y 2 . This function is also continuous except along the line y = 0, and so it is continuous in a neighborhood of (1 , 1). We can then conclude, via the Existence and Uniqueness Theorem, that the given IVP has a unique solution. 5 Make an appropriate substitution in order to solve the following differential equations. a. dy dx = 2 y x x 2 y 2 .

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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 8 Solution: This equation is of the form a ( x ) dy dx = b ( x ) y + c ( x ) y n (here n = 2), so it is Bernoulli. To solve a Bernoulli equation we divide by the greatest power of y and then let our substitution v equal the second smallest power of y in the remaining equation. So in this case we have y - 2 dy dx = 2 x y - 1 x 2 . We set v = y - 1 , and then have v = y - 2 dy dx by the Chain Rule. Therefore we have dv dx = 2 x v x 2 . This is a linear equation (all Bernoulli equations transform into linear equations), so we put it in standard form, dv dx + 2 x v = x 2 . Our integrating factor is then μ ( x ) = e integraltext 2 x dx = e 2ln | x | = e | x | 2 = | x | 2 = x 2 . After multiplying through by x 2 , we have x 2 dv dx + 2 xv = x 4 . We now integrate both sides by x . We know that the left-hand side will become μ ( x ) v , while the right-hand side is just x 5 / 5 + C , so x 2 v = x 5 5 + C, v = x 3 5 + C x 2 , 1 y = x 3 5 + C x 2 , y ( x ) = 1 x 3 5 + C x 2 .
MAC 2313, Fall 2010 — Midterm 1 Review Solutions 9 b. x 2 dy dx = xy y 2 . Solution: This is a homogeneous equation 3 , that is, an equation of the form dy dx = G ( y / x ), so we make the substitution v = y / x . First we need to translate the dy dx into these terms: y = xv, so dy dx = x dv dx + v by the product rule. Maying this substitution, we transform the equation into x dv dx + v = v v 2 , which separates (after canceling a v on each side) as v - 2 dv dx = x - 1 . Integrating both sides, we obtain integraldisplay v - 2 dv dx dx = integraldisplay x - 1 dx, v - 1 = ln | x | + C, v = 1 ln | x | + C . Finally we put this back in terms of y ( x ): y x = 1 ln | x | + C , so y ( x ) = x ln | x | + C . c. dy dx = 1 (2 x + y ) e 2 x + y 2. 3 This equation is also a Bernoulli equation, so you could solve it by dividing by y 2 and then setting v = y - 1 .

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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 10 Solution: This is an equation of the form dy dx = G ( ax + by ), so we make the substitution v = 2 x + y . We then see that dv
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