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

Info icon This preview shows pages 1–6. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(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
Image of page 2
(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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Image of page 4
(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.
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern