MATH
Prep

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

• Notes
• 95
• 100% (1) 1 out of 1 people found this document helpful

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

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

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

(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
(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

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

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
(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.

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

This is the end of the preview. Sign up to access the rest of the document.
• Winter '09
• SidneyTrudeau
• Math, Equations, Quadratic equation, Trigraph, Characteristic polynomial, dx, yp

{[ snackBarMessage ]}

### What students are saying

• 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.

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

• 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.

Dana University of Pennsylvania ‘17, Course Hero Intern

• 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.

Jill Tulane University ‘16, Course Hero Intern