Prep - 0.1 First Order Dierential Equations(a y = e y We...

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0.1 First Order Differential Equations 1. Solve the following separable differential equations. (a) y 0 = e - 2 x y . We start by separating the equation: dy dx = e - 2 x y y dy = e - 2 x dx Next, we integrate both sides: Z y dy = Z e - 2 x dx y 2 2 = - e - 2 x 2 + C Lastly, solving for y : y 2 2 = - e - 2 x 2 + C y 2 = - e - 2 x + 2 C y = ± p - e - 2 x + 2 C = ± p - e - 2 x + D D = 2 C. (b) 3 y 5 y 0 - x 2 ( x 3 + 5) 10 = 0 We start by separating the equation: 3 y 5 dy dx = x 2 ( x 3 + 5) 10 3 y 5 dy = x 2 ( x 3 + 5) 10 dx Next, we integrate both sides: Z 3 y 5 dy = Z x 2 ( x 3 + 5) 10 dx Let u = x 3 + 5, so du = 3 x 2 dx : y 6 2 = Z 1 3 u 10 du = 1 33 u 11 + C = 1 33 ( x 3 + 5) 11 + C Lastly, solving for y : y 6 2 = 1 33 ( x 3 + 5) 11 + C y 6 = 2 33 ( x 3 + 5) 11 + 2 C = 2 33 ( x 3 + 5) 11 + D D = 2 C. 1
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(c) y 0 = e x - 2 y +2 . We start by separating the equation: dy dx = e x - 2 y + 2 ( y + 2) dy = e x - 2 dx Next, we integrate both sides: Z ( y + 2) dy = Z e x - 2 dx y 2 2 + 2 y = e x - 2 x + C Lastly, we solve for y using the quadratic formula: y 2 2 + 2 y - ( e x - 2 x + C ) = 0 y = - 2 ± p 4 - 2( e x - 2 x + C ) = - 2 ± 4 - 2 e x - 4 x + D D = 2 C. (d) y 0 = - 1+2 y x (1+ y ) . We start by separating the equation: dy dx = - 1 + 2 y x (1 + y ) 1 + y 1 + 2 y dy = - 1 x dx Next, we integrate both sides: Z 1 + y 1 + 2 y dy = - Z 1 x dx Z 1 2 (1 + 2 y ) + 1 2 1 + 2 y dy = - Z 1 x dx Z 1 2 + 1 2 1 1 + 2 y dy = - Z 1 x dx 1 2 y + 1 4 ln | 1 + 2 y | = - ln x + C Here, we cannot solve for y , so we leave the solution in implicit form. 2
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(e) y 0 + x 2 y = x 2 y 2 . We start by separating the equation: dy dx = x 2 y 2 - x 2 y = x 2 ( y 2 - y ) dy y 2 - y = x 2 dx Next, we integrate both sides: Z dy y 2 - y = Z x 2 dx. We solve the left hand side by partial fractions: 1 y 2 - y = - 1 y + 1 y - 1 , so - ln | y | + ln | y - 1 | = x 3 3 + C ln | y - 1 y | = x 3 3 + C ln | 1 - 1 y | = x 3 3 + C Lastly, we solve for y : | 1 - 1 y | = e x 3 3 + C = De x 3 3 D = e C 1 - 1 y = ± De x 3 3 y = 1 1 ± De x 3 3 3
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2. Solve the following homogeneous equations. (a) y 0 = y 5 - x 5 y 4 x . Start by checking that the equation is homogeneous: let f ( x, y ) = y 5 - x 5 y 4 x . Then, f ( xt, yt ) = ( yt ) 5 - ( xt ) 5 ( yt ) 4 ( xt ) = t 5 ( y 5 - x 5 ) t 5 ( y 4 x ) = y 5 - x 5 y 4 x = f ( x, y ) . Let y = vx , so dy dx = v + x dv dx . Substitute these into the equation: v + x dv dx = (( vx ) 5 - x 5 ( vx ) 4 x = x 5 ( v 5 - 1) x 5 v 4 = v 5 - 1 v 4 x dv dx = v 5 - 1 v 4 - v = - 1 v 4 . The new equation is separable: v 4 dv = - dx x Z v 4 dv = - Z dx x v 5 5 = - ln | x | + C Let C = - ln | k | , so v 5 5 = - ln | x | - ln | k | = - ln | kx | v 5 = - 5 ln | kx | = ln ( | kx | - 5 ) v = 5ln ( | kx | - 5 ) Lastly, writing the expression in terms of y gives: y = vx = x 5ln ( | kx | - 5 ) . 4
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(b) y 0 = - 2 y +3 x y . Start by checking that the equation is homogeneous: let f ( x, y ) = - 2 y +3 x y . Then, f ( xt, yt ) = - 2( yt ) + 3( xt ) ( yt ) = - 2 y + 3 x y = f ( x, y ) . Let y = vx , so dy dx = v + x dv dx . Substitute these into the equation: v + x dv dx = - 2( vx ) + 3 x vx = - 2 v + 3 v x dv dx = - 2 v + 3 v - v = - v 2 - 2 v + 3 v v v 2 + 2 v - 3 dv = - dx x Z v v 2 + 2 v - 3 dv = - Z dx x The right hand side can be done by partial fractions: v v 2 + 2 v - 3 = 3 4 1 v + 3 + 1 4 1 v - 1 , so Z 3 4 1 v + 3 + 1 4 1 v - 1 dv = - Z dx x 3 4 ln | v + 3 | + 1 4 ln | v - 1 | = - ln | x | + C Let C = - ln | k | , so 3 4 ln | v + 3 | + 1 4 ln | v - 1 | = - ln | x | - ln | k | = - ln | kx | 3 ln | v + 3 | + ln | v - 1 | = - 4 ln | kx | ln | v + 3 | 3 + ln | v - 1 | = ln | kx | - 4 ln | v + 3 | 3 | v - 1 | = ln | kx | - 4 | v + 3 | 3 | v - 1 | = | kx | - 4 | y x + 3 | 3 | y x - 1 | = | kx | - 4 This expression cannot be simplified further, so we leave it as is.
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  • Winter '09
  • SidneyTrudeau
  • Math, Equations, Quadratic equation, Trigraph, Characteristic polynomial, dx, yp

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