<p>"Consider the differential equation y'=xy, with initial condition y(0)=1. Using Euler's method starting at x=0 with step size 1, what is the approximate value of y(2)?"</p>
<p>What I did: </p>
<p>x1= 1
y'1= 1(1)=1
y1=1+1(1)=2</p>
<p>x2=2
y'2=2(2)=4
y2=2+4(1)=6</p>
<p>so therefore y(2) = 6, right?</p>
<p>But the Kaplan's book says that y(2) is 2 and I don't know how. Did I do something wrong?</p>
<p>When x = 1, y = 1 (since dy when x = 0 is 0) and dy = 1 so when x = 2, y(2) = 1 + 1 = 2.</p>
<p>Basically you were right at x1 = 1, y’1 = 1,
but you wrote y1 = 1+1 = 2, when its actually y2 = 1+1 = 2. </p>
<p>y(3) is 6.</p>
<p>Thank you; I appreciate your help!</p>
<p>By the way, if you get this type of question on the Calculator section, you can usually separate and solve for the exact value and it should be reasonably close. </p>
<p>If you are lucky, two answer choices will fill around the exact, one will be over and one will be under. Drawing a quick graph and thinking about concavity should tell if you the linearization of Euler’s method will over or under approximate it.</p>