Euler Equations — Question 1

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Question 1

Consider the Euler initial-value problem x2y″−2xy′−4y=0,y(1)=2,y′(1)=3.x^2y''-2xy'-4y=0,\qquad y(1)=2,\quad y'(1)=3. A power xmx^m is a useful trial function, but the reason it works should also be understood.

Tasks

  1. For x>0x>0, set t=ln⁡xt=\ln x and Y(t)=y(et)Y(t)=y(e^t). Derive the formulas for xy′xy' and x2y″x^2y'' and transform the equation into one with constant coefficients.

  2. Find a fundamental pair, compute its Wronskian, and solve the IVP exactly. Verify both initial values.

  3. State the maximal real interval of this IVP and determine the behavior of its solution at each end of that interval.

  4. Locate and classify every stationary point on that interval. Give the exact minimum value and explain how two monotone power modes can combine into a nonmonotone solution.

Original worksheet page 1: question and worked solution for 6-4-001
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Question 1 – Solution

Strategy. Logarithmic coordinates turn powers into exponentials and make the characteristic equation systematic.

Step 1: Transform derivatives correctly. The chain rule gives Y′=xy′Y'=xy' and Y″=xy′+x2y″Y''=xy'+x^2y'', hence xy′=Y′,x2y″=Y″−Y′.\boxed{xy'=Y',\qquad x^2y''=Y''-Y'.} The equation becomes Y″−3Y′−4Y=0Y''-3Y'-4Y=0, with characteristic polynomial m2−3m−4=(m−4)(m+1)m^2-3m-4=(m-4)(m+1).

Step 2: Solve and verify the data. The powers x4,x−1x^4,x^{-1} form a pair with W=x4(−x−2)−4x3x−1=−5x2≠0W=x^4(-x^{-2})-4x^3x^{-1}=-5x^2\ne 0 for x>0x>0. Writing y=C1x4+C2x−1y=C_1x^4+C_2x^{-1} gives C1+C2=2C_1+C_2=2, 4C1−C2=34C_1-C_2=3, so y=x4+x−1.\boxed{y=x^4+x^{-1}.} At 11, its value is 22 and derivative 4x3−x−24x^3-x^{-2} is 33.

Step 3: Identify the maximal interval. The normalized coefficients are continuous on (0,∞)(0,\infty), and the explicit solution exists throughout it. Since y→+∞y\to+\infty as x→0+x\to 0^+, no finite continuation through 00 is possible. Thus the maximal real interval through 11 is (0,∞)\boxed{(0,\infty)}. At infinity, y∼x4→+∞y\sim x^4\to+\infty; at zero, y∼x−1y\sim x^{-1}.

Step 4: Find the unique minimum. The equation y′=0y'=0 is 4x5=14x^5=1, giving x*=4−1/5x_*=4^{-1/5}. Since y″=12x2+2x−3>0y''=12x^2+2x^{-3}>0, it is the unique global minimum. Using x*−1=4x*4x_*^{-1}=4x_*^4, y(x*)=5⋅4−4/5.\boxed{y(x_*)=5\cdot 4^{-4/5}.} One mode increases and the other decreases. Their derivatives cancel at exactly one point, although neither mode has a stationary point of its own.

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Original worksheet page 2: question and worked solution for 6-4-001

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