2302-practice-mid1-soln

# Equation is linear however because we can express it

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equation is linear, however, because we can express it as dy dx + 1 2 x y = 1 2 . The equation is not exact, because if we express it as M ( x, y ) + N ( x, y ) dy dx = 0, we see that M ( x, y ) = y 2 x 1 2 and N ( x, y ) = 1 , and the partials do not match: ∂M ∂y = 1 2 x negationslash = 0 = ∂N ∂x . Therefore we can only solve this equation by use of an integrating factor. Recall that this integrating factor is e integraltext P ( x ) dx when the equation is put in the form dy dx + P ( x ) y = Q ( x ): e integraltext P ( x ) dx = e integraltext 1 2 x dx = e 1 2 integraltext 1 x dx = e ln x 1 / 2 = x 1 / 2 . Multiplying through by the integrating factor, we obtain x 1 / 2 dy dx + 1 2 x 1 / 2 y = 1 2 x 1 / 2 . Now we integrate both sides with respect to x . We know that the left-hand side of this equation is the derivative (with respect to x ) of the integrating factor times y : x 1 / 2 y . On the right-hand side, we have integraldisplay 1 x 1 / 2 dx = 1 2 parenleftbigg 2 3 x 3 / 2 parenrightbigg + C = 1 3 x 3 / 2 + C, so we obtain x 1 / 2 y = 1 3 x 3 / 2 + C, and thus y ( x ) = 1 3 x + C x 1 / 2 . It is important to note that the term on the right of this solution is a constant times a function of x , and therefore we cannot replace this term by a simple constant. e. dy dx = 5 x 4 cos y + e y .

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MAC 2313, Fall 2010 — Midterm 1 Review Solutions 6 Solution: This equation is separable, because we can express it as (cos y + e y ) dy dx = 5 x 4 . Therefore the equation is also exact. The equation is not linear, however, because e y term prevents us from expressing it in the form dy dx + P ( x ) y = Q ( x ). To solve the equation, we integrate both sides of its separated form above (with respect to x ): integraldisplay cos y + e y dy dx dx = integraldisplay 5 x 4 dx, integraldisplay cos y + e y dy = x 5 + C, sin y + e y = x 5 + C. We have no way to solve this equation for y ( x ), so we have to be satisfied with this implicit solution. 4 For each of the following initial value problems, determine if they have zero, one, or more than one solution(s). You do not need to solve these equations. a. y dy dx + x = 0; y (1) = 0. Solution: This equation has no solutions. Plugging in the initial conditions ( x 0 , y 0 ) = (1 , 0) gives (0) dy dx (1) + 1 = 0 , i.e., 1 = 0, which has no solution. b. dy dx = 3 y 2 / 3 ; y (0) = 0. Solution: This solution has more than one solution . In particular, y 0 ( y identically equal to 0 solves the differential equation, and y ( x ) = x 3 also solves the equation (this solution can
MAC 2313, Fall 2010 — Midterm 1 Review Solutions 7 be found by separating the equation). In fact, this equation has infinitely many solutions : for every number a , the function y ( x ) = braceleftbigg 0 for x a , ( x a ) 3 for x > a solves the equation.

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