Differential Equations Question - None

Question

Let $\frac{x\,dy}{dx} - y = x^2\left(xe^x + e^x - 1\right)$ for all $x \in \mathbb{R} - \{0\}$ such that $y(1) = e - 1$. If $y(2) = k\,y(1)\,(y(1) + 2)$, then the value of $\dfrac{k^2}{5}$ is

Accepted Answer

Step 1: Rewrite the equation in standard form by dividing by x:

dy/dx - y/x = x²e^x + xe^x - x

Step 2: Identify as linear ODE with P(x) = -1/x. The integrating factor is:

μ(x) = e^(-∫1/x dx) = e^(-ln|x|) = 1/x

Step 3: Multiply both sides by μ(x) = 1/x:

(1/x)·dy/dx - y/x² = xe^x + e^x - 1

Step 4: Recognize left side as d/dx(y/x):

d/dx(y/x) = xe^x + e^x - 1

Step 5: Integrate both sides:

y/x = ∫(xe^x + e^x - 1)dx

Using integration by parts on xe^x: ∫xe^x dx = xe^x - e^x

y/x = (xe^x - e^x) + e^x - x + C = xe^x - x + C

Step 6: Apply initial condition y(1) = e - 1:

(e - 1)/1 = 1·e - 1 + C

e - 1 = e - 1 + C ⟹ C = 0

Step 7: Therefore y = x(xe^x - x) = x²e^x - x²

Step 8: Find y(2):

y(2) = 4e² - 4

Step 9: Use condition y(2) = k·y(1)·(y(1) + 2):

4e² - 4 = k(e - 1)(e - 1 + 2)

4(e² - 1) = k(e - 1)(e + 1)

4(e - 1)(e + 1) = k(e - 1)(e + 1)

k = 4

Step 10: Calculate k²/5:

k²/5 = 16/5

∴ Answer: 16/5 or 3.2

Question:

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