This preview shows page 1. Sign up to view the full content.
Unformatted text preview: CHAPTER 2: First Order Diﬀerential Equations
EXERCISES 2.2: Separable Equations, page 46 1. This equation is separable because we can separate variables by multiplying both sides by dx and dividing by 2y 3 + y + 4. 3. This equation is separable because yex+y dy =2 = dx x +2 5. Writing the equation in the form ds s+1 = − s2 , dt st we see that the right-hand side cannot be represented in the form g (t)p(s). Therefore, the equation is not separable. 7. Multiplying both sides of the equation by y 2 dx and integrating yields y 2dy = (1 − x2 )dx ⇒ ⇒ ⇒ y 2 dy = (1 − x2 )dx ⇒ y= √ 3 3x − x3 + C , ex x2 + 2 y ey = g (x)p(y ). 13 1 y = x − x3 + C1 3 3 y 3 = 3x − x3 + C where C := 3C1 is an arbitrary constant. 9. To separate variables, we divide the equation by y and multiply by dx. This results dy dy = y (2 + sin x) ⇒ = (2 + sin x)dx dx y dy = (2 + sin x)dx ⇒ ln |y | = 2x − cos x + C1 ⇒ y ⇒ |y | = e2x−cos x+C1 = eC1 e2x−cos x = C2 e2x−cos x , 27 ...
View Full Document