AP Calculus BC 7.6 Finding General Solutions Using Separation of Variables - FRQs - Exam Style Questions
No-Calc Question
Let \(y=f(x)\) be the particular solution to the differential equation \[ \frac{dy}{dx}=y\cdot\big(x\ln x\big), \] with initial condition \(f(1)=4\). It is given that \(f”(1)=4\).
(a) Write the second-degree Taylor polynomial for \(f\) about \(x=1\). Then use it to approximate \(f(2)\).
(b) Use Euler’s method, starting at \(x=1\) with two steps of equal size, to approximate \(f(2)\). Show your work.
(c) Find the particular solution \(y=f(x)\) to \(\displaystyle \frac{dy}{dx}=y\cdot(x\ln x)\) with \(f(1)=4\).
Most-appropriate topic codes (CED):
• TOPIC 10.11: Finding Taylor Polynomial Approximations of Functions — part (a)
• TOPIC 7.5: Approximating Solutions Using Euler’s Method — part (b)
• TOPIC 7.6: Separation of Variables (general solutions) — part (c)
• TOPIC 7.7: Particular Solutions Using Initial Conditions — part (c)
• TOPIC 7.5: Approximating Solutions Using Euler’s Method — part (b)
• TOPIC 7.6: Separation of Variables (general solutions) — part (c)
• TOPIC 7.7: Particular Solutions Using Initial Conditions — part (c)
▶️ Answer/Explanation
(a) Second-degree Taylor polynomial at \(x=1\)
From \(\displaystyle \frac{dy}{dx}=y(x\ln x)\) and \(f(1)=4\): \(f'(1)=4\cdot(1\cdot\ln 1)=0\). Given \(f”(1)=4\).
\(\displaystyle T_{2}(x)=f(1)+f'(1)(x-1)+\frac{f”(1)}{2}(x-1)^{2} =4+0\cdot(x-1)+\frac{4}{2}(x-1)^{2}=4+2(x-1)^{2}.\)
\(\displaystyle f(2)\approx T_{2}(2)=4+2(1)^{2}=6.\)
(b) Euler’s method, two steps from \(x=1\) to \(x=2\) (\(h=0.5\))
Slope formula: \(\displaystyle y’=y(x\ln x)\). Start \((x_{0},y_{0})=(1,4)\).
Step 1 to \(x_{1}=1.5\): slope \(=4\cdot(1\cdot\ln 1)=0\) ⇒ \(y_{1}\approx y_{0}+h\cdot 0=4\).
Step 2 to \(x_{2}=2\): slope at \((1.5,4)\) is \(4\cdot\big(1.5\ln 1.5\big)\).
\(\displaystyle y_{2}\approx y_{1}+h\cdot 4(1.5\ln 1.5)=4+\tfrac12\cdot 6\ln 1.5 =4+3\ln 1.5.\)
So \(\displaystyle f(2)\approx 4+3\ln 1.5\) (≈ \(4.? + \) numeric if desired).
(c) Particular solution
Separate variables: \(\displaystyle \frac{1}{y}\,dy=x\ln x\,dx\).
Integrate: \(\displaystyle \int \frac{1}{y}\,dy=\int x\ln x\,dx\).
By parts (\(u=\ln x,\ dv=x\,dx\)): \(\displaystyle \int x\ln x\,dx=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C\).
Thus \(\displaystyle \ln|y|=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C\).
Apply \(f(1)=4\): \(\displaystyle \ln 4=0-\frac14+C \Rightarrow C=\ln 4+\frac14.\)
Hence \[ y=f(x)=\exp\!\Big(\tfrac{x^{2}}{2}\ln x-\tfrac{x^{2}}{4}+\ln 4+\tfrac14\Big), \qquad (x>0). \] (Equivalently \( \displaystyle y=4e^{1/4}\,x^{\,x^{2}/2}\,e^{-x^{2}/4}\,. \))
From \(\displaystyle \frac{dy}{dx}=y(x\ln x)\) and \(f(1)=4\): \(f'(1)=4\cdot(1\cdot\ln 1)=0\). Given \(f”(1)=4\).
\(\displaystyle T_{2}(x)=f(1)+f'(1)(x-1)+\frac{f”(1)}{2}(x-1)^{2} =4+0\cdot(x-1)+\frac{4}{2}(x-1)^{2}=4+2(x-1)^{2}.\)
\(\displaystyle f(2)\approx T_{2}(2)=4+2(1)^{2}=6.\)
(b) Euler’s method, two steps from \(x=1\) to \(x=2\) (\(h=0.5\))
Slope formula: \(\displaystyle y’=y(x\ln x)\). Start \((x_{0},y_{0})=(1,4)\).
Step 1 to \(x_{1}=1.5\): slope \(=4\cdot(1\cdot\ln 1)=0\) ⇒ \(y_{1}\approx y_{0}+h\cdot 0=4\).
Step 2 to \(x_{2}=2\): slope at \((1.5,4)\) is \(4\cdot\big(1.5\ln 1.5\big)\).
\(\displaystyle y_{2}\approx y_{1}+h\cdot 4(1.5\ln 1.5)=4+\tfrac12\cdot 6\ln 1.5 =4+3\ln 1.5.\)
So \(\displaystyle f(2)\approx 4+3\ln 1.5\) (≈ \(4.? + \) numeric if desired).
(c) Particular solution
Separate variables: \(\displaystyle \frac{1}{y}\,dy=x\ln x\,dx\).
Integrate: \(\displaystyle \int \frac{1}{y}\,dy=\int x\ln x\,dx\).
By parts (\(u=\ln x,\ dv=x\,dx\)): \(\displaystyle \int x\ln x\,dx=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C\).
Thus \(\displaystyle \ln|y|=\frac{x^{2}}{2}\ln x-\frac{x^{2}}{4}+C\).
Apply \(f(1)=4\): \(\displaystyle \ln 4=0-\frac14+C \Rightarrow C=\ln 4+\frac14.\)
Hence \[ y=f(x)=\exp\!\Big(\tfrac{x^{2}}{2}\ln x-\tfrac{x^{2}}{4}+\ln 4+\tfrac14\Big), \qquad (x>0). \] (Equivalently \( \displaystyle y=4e^{1/4}\,x^{\,x^{2}/2}\,e^{-x^{2}/4}\,. \))