Math 201 Solutions to Assignment 1. 2ydy = x 2 dx. y = C 1 3 x3
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1 Math 201 Solutions to Assignment 1 1. Solve the initial value problem: x 2 dx + 2y = 0, y(0) = 2. x 2 dx + 2y = 0, y(0) = 2 2y = x 2 dx y 2 = 1 3 x3 + C y = C 1 3 x3 Notice that y is not defined for some values of x. Applying the initial condition gives us that y(0) = C = 2, so C = 4. Then y = x3. 2. Find the general solution to the equation: The integrating factor is given by: x dx + 3(y + x2 ) = sin x x. µ (x) = 3µ x and after integration we find that µ(x) = x 3. Multiply the standard form of the equation by µ to obtain (x 3 y) = 3x 4 + x sin x 1
2 Integrating both sides we get x 3 y = 3/5x 5 + x sin xdx. integrating the integral above by parts and dividing the equation by x 3 we obtain y = 3/5x 2 cos x/x 2 + sin x/x 3 + C/x Solve the initial value problems: (ye xy 1/y)dx + (xe xy + x/y 2 ) = 0, y(1) = 1, Using the test for exactness one verifies that the equation is exact. Then F/ x = ye xy 1/y and after integration Differentiating with respect to y we get On the other hand we need that F (x, y) = e xy x/y + g(y) (1) F/ y = xe xy + x/y 2 + g (y) F/ y = xe xy + x/y 2 and therefore g (y) = 0. Thus we can choose g(y) = 0 because any constant can be absorbed in the right hand side constant of the solution which we easily obtain from (1) to be: e xy x y = const = C Since we want y(1) = 1 it follows that F (1, 1) = e 1 = C 2
3 and then the solution of the initial-value problem (IVP) is given by e xy x y = e 1. (e t x + 1)dt + (e t 1)dx = 0, x(1) = 1. Again we can verify that this is an exact equation. Then F/ t = e t x + 1 and integrating we obtain F (t, x) = xe t + t + g(x) Differentiating with respect to y F/ x = e t + g (x) On the other hand or g = x F/ x = e t 1 g (x) = 1 F (t, x) = xe t + t x = const = C is a solution. We need that x(1) = 1 or F (1, 1) = e = C which gives the unique solution to the IVP x = e t e t 1 3
4 4. Find the general solution to the equations: dx y(ln y ln x + 1) =. x Note that x and y must be both positive for the logarithms to be well defined. The equation can be put in the form dx = y (ln y ) x x + 1. This is a homogeneous equation and we need to substitute u = y/x. Then the equation in terms of u reads x du + u = u(ln u + 1). dx This is a separable equation and the separated form is After one integration we obtain du u ln u = dx x. ln ln u = ln x + c 1 and taking the exponents of both sides ln u = cx. This yields y = xe cx. dx y = e2x y 3 4
5 We first divide by y 3 and obtain 3 y dx y 2 = e 2x. Note that while dividing by y we presume that y 0 but y = 0 is a solution to the initial equation which will not be a solution to the new equation. Therefore, we have to keep it in mind. We further substitute y 2 = u and after some simple calculus (du/dx = 2y 3 /dx) obtain du dx + 2u = 2e2x. This is a linear equation and the integrating factor is µ(x) = e 2x. Multiplying the equation with it we obtain d dx (e2x u) = 2e 4x and integrating once, and substituting back u = y 2 y 2 = 1 2 e2x + ce 2x. So the solutions are y 2 = 1 2 e2x + ce 2x and y = 0. dx = (2x + y 1)2, The equation is of the type: G(ax + by) with a = 2, b = 1. substitute v = 2x + y to obtain: We dx = 2 + dx and therefore the equation becomes dx = (v 1)
6 which is a separable equation. Then which after one integration yields or Substituting v = 2x + y we get (v 1) = dx 1 2 arctan v 1 2 = x + c 1 v = 2 tan( 2x + c), c = 2c 1 y = 1 2x + 2 tan( 2x + c) dθ + y θ = 4θy 2, The equation is of a Bernoulli type with n = 2. Therefore we substitute v = y 3 which yields dθ The equation transforms into or (substituting y 3 = v) = 1/3y 2 dθ 2 1/3y dθ + y θ = 4θy 2 dθ + 3 θ v = 12θ which is a linear equation. µ(x) = θ 3 and multiplying by it we get d dθ [θ3 v] = 12θ 4 6
7 which after one integration gives v = 12/5θ 2 + c/θ 3 or y = ( 12/5θ 2 + c/θ 3 ) 1/3 dx = x y 1 x + y + 5 Some elementary algebra (as shown in p. 73 of the textbook, 8th edition) shows that the substitution x = u 2, y = v 3 reduces the equation to the homegeneous one: du = u v u + v Therefore we substitute v/u = w which yields udu v u = dw 2 du Then dw du u = w + 1 w 1 + w After some elementary calculations we obtain Integrating once both sides we get 1 + w 1 2w w 2 dw = du u 1 2 dp 2 p = ln u + c where p = (w + 1) 2. After computing the integral we get 1/2 ln (w + 1) 2 2 = ln u + c 7
8 or after taking the exponent of both sides (w + 1) 2 2 = Au 2. After substituting back w, u and v and some elementary algebra we obtain (y + 3) 2 + 2(x + 2)(y + 3) (x + 2) 2 = A or 5. Solve the initial value problem: y 2 x 2 + 2xy + 10y + 2x = C ( x dx = y + y ), y(1) = 4. x The equation is obviously homogeneous and therefore we substitute v = y x which yields dx = x dx + v After substitution in the equation we get x dx = 1 v which is a separable equation. After integrating it we obtain the following solution 1/2v 2 = ln x + c or after substituting v = y/x y 2 = 2 ln x + c x2 To obtain the solution of the IVP we impose that y 2 (1) = 16 which yields c = 16. 8
for any C, including C = 0, because y = 0 is also a solution: dy
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