Question
Let y = f(x) be the particular solution to the differential equation \(\frac{dy}{dx}=y\cdot (x In x)\) with initial condition f(1) = 4 . It can be shown that f”(1) = 4.
(a) Write the second-degree Taylor polynomial for f about x = 1. Use the Taylor polynomial to approximate f(2) .
(b) Use Euler’s method, starting at x = 1 with two steps of equal size, to approximate f(2) . Show the work that leads to your answer.
(c) Find the particular solution y = f(x) to the differential equation \(\frac{dy}{dx}=y\cdot (x In x)\) with initial condition f(1) = 4 .
Answer/Explanation
Ans:
(a)
\(\frac{dy}{dx}= 4.1 In 1\)
\(P_{2}(x)=4+In 1 (x-1)+\frac{4(x-1)^{2}}{2!}\) \(P_{2}(x)=4+\frac{4(x-1)^{2}}{2!}\)
\(P_{2}(2)=4+\frac{4(2-1)^{2}}{2!}= 4+2 = 6\)
(b)
\(\frac{dy}{dx}= 4\cdot 1.5 In 1.5\)
6 In 1.5
f(2) = 4 + 3 In 1.5
(c)
\(\frac{dy}{dx}= y\cdot sin x\)
\(\frac{1}{y}dy= x In x dx\) u = In x dv = x dx
\(\int \frac{1}{y}dy= \int x In x dx\) \(du = \frac{1}{x}dx\) \(v = \frac{1}{x}x^{2}\)
\(In y = \frac{1}{2}x^{2}In x -\frac{1}{4}x^{2}+ c\)
\(In 4 = \frac{1}{2}In 1 -\frac{1}{4}+ c\)
\(In 4 + \frac{1}{4}= c\)
\(In y =\frac{1}{2}x^{2}In x- \frac{1}{4}x^{2}+In 4 + \frac{1}{4}\)
\(y =4e^{\frac{1}{2}x^{2}In x}- \frac{1}{4}x^{2} + \frac{1}{4}\)