Exam 2: First-Order Differential Equations

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Solve the problem y=(x+1)y,y(0)=1y ^ { \prime } = ( x + 1 ) y , y ( 0 ) = 1 numerically for y(0.2)y ( 0.2 ) using h=0.1h = 0.1 .

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The autonomous differential equation dxdt=x2(x4)\frac { d x } { d t } = x ^ { 2 } ( x - 4 ) has a solution that is

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The solution of (x2y)dx+ydy=0( x - 2 y ) d x + y d y = 0 is

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The differential equation 2xydx+(x2+1)dy=02 x y d x + \left( x ^ { 2 } + 1 \right) d y = 0 is

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If a>0a > 0 and b>0b > 0 , the autonomous differential equation dPdt=P(abP)\frac { d P } { d t } = P ( a - b P ) has a solution that is

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The differential equation (xysinx+2ycosx)dx+2xcosxdy=0( - x y \sin x + 2 y \cos x ) d x + 2 x \cos x d y = 0 is

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The differential equation y=(4x+2y+3)2y ^ { \prime } = ( 4 x + 2 y + 3 ) ^ { 2 } has the solution

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The solution of the differential equation y=x2yy ^ { \prime } = x ^ { 2 } y is

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The solution of the differential equation y+y/x=y2y ^ { \prime } + y / x = y ^ { 2 } is

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The differential equation (y3+6xy4)dx+(3xy2+12x2y3)dy=0\left( y ^ { 3 } + 6 x y ^ { 4 } \right) d x + \left( 3 x y ^ { 2 } + 12 x ^ { 2 } y ^ { 3 } \right) d y = 0 is

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The solution of the differential equation y=xyy ^ { \prime } = x y is

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The solution of (x+2y)dx+ydy=0( x + 2 y ) d x + y d y = 0 is

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The differential equation y+y/x=y2y ^ { \prime } + y / x = y ^ { 2 } can be solved using the substitution

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The differential equation (x+2y)dx+ydy=0( x + 2 y ) d x + y d y = 0 can be solved using the substitution

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The differential equation yy/x=y2y ^ { \prime } - y / x = y ^ { 2 } can be solved using the substitution

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An integrating factor for the linear differential equation x2y+xy=1x ^ { 2 } y ^ { \prime } + x y = 1 is

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The differential equation xy=2y+sinxx y ^ { \prime } = 2 y + \sin x is

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The differential equation 2xydx+(x2+y3)dy=02 x y d x + \left( x ^ { 2 } + y ^ { 3 } \right) d y = 0 is

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The differential equation y=x+y+11y ^ { \prime } = \sqrt { x + y + 1 } - 1 has the solution

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The solution of the differential equation yy=xy ^ { \prime } - y = x is

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