578 CHAPTER 29. FIRST ORDER SCALAR ODE

75. Solve the following initial value problems and give a graph of each on [0,3] on thesame axes. y′= 1

10 y(5− y) ,y(0)= .3,y′= 110 y(5− y) ,y(0)= .5,y′= 1

10 y(5− y) ,y(0)=−.3.

76. Give a graph of the solutions to the differential equation y′ = ty2− (.1)y3 on theinterval [0,5] which result from the initial conditions y(0) = 1,0,2,−3.

77. Give a graph of the solution to y′ = x(y2)3/4− xy3 +1, y(0) = 0.

78. Use a computer algebra system to obtain a solution to the initial value problem

y′ =y3

x3 +8y3 , y(0) = 1

You may have to obtain a numerical solution in terms of a graph. It is true that theequation is homogeneous, but it might be too hard to carry out the computations.Scientific notebook has trouble with this one.

79. Use a computer algebra system to obtain the graph of the solution to the initial valueproblem x2y′ = 4x2 + xy+ y2,y(4) = 1.

80. Find the solution to the following initial value problem, either a graph or a formula.Then graph it

y′ = xy+ sin(x)− 110

y2, y(0) = 1

81. When you use MATLAB or other computer algebra system to find a numerical solu-tion to a differential equation, you are using a fairly sophisticated numerical method.The most primitive method for obtaining numerical solutions to y′ = f (t,y) is calledEuler’s method. In this method, one has a step size h and partitions the time intervalinto t0 < t1 < · · · < tn = T, t j+1 = t j + h. Then letting y0 be the initial condition,Euler’s method goes like this. You iterate the following process.

k = f (ti,yi) , yi+1 = yi +hk, ti+1 = ti +h

When you get to tn, you stop. Your solution consists of a function y which interpo-lates the points (ti,yi) meaning yi = y(ti). You can easily get MATLAB to do this foryou. Here is the case of y′ = y,y(0) = 1.

f=@(t,y) y; h=.01; y(1)=1; t(1)=0;hold on; for j=1:500;k=f(t(j),y(j)); y(j+1)=y(j)+h*k; t(j+1)=t(j)+h;end; plot(t,y); [t(501),y(501)]

The first line is defining the function f (t,y) = y. Thus the real solution is et . Thenumber y(501) is the Euler solution at 5. Compare with e5.

82. Suppose you have the initial value problem y′ = y,y(0) = y0. You know the solutionis ety0. Consider the interval [0, t]. Consider for k ≤ n

yk+1 = yk +tn

yk