Differential Equations Question - None

Question

For $x \in \mathbb{R},\, x \neq 0$, if $y(x)$ is a differentiable function such that $x\int_1^x y(t)\,dt = (x+1)\int_1^x ty(t)\,dt$, then $y(x)$ equals (where $C$ is a constant)

Accepted Answer

Step 1: Given: $x\int_1^x y(t)\,dt = (x+1)\int_1^x ty(t)\,dt$

Step 2: Differentiate both sides with respect to $x$ using product rule and Leibniz rule:

Left side: $\frac{d}{dx}\left[x\int_1^x y(t)\,dt\right] = \int_1^x y(t)\,dt + x\cdot y(x)$

Right side: $\frac{d}{dx}\left[(x+1)\int_1^x ty(t)\,dt\right] = \int_1^x ty(t)\,dt + (x+1)\cdot xy(x)$

Step 3: Equating both sides:

$\int_1^x y(t)\,dt + xy(x) = \int_1^x ty(t)\,dt + (x+1)xy(x)$

Step 4: Rearranging:

$\int_1^x y(t)\,dt - \int_1^x ty(t)\,dt = (x+1)xy(x) - xy(x)$

$\int_1^x y(t)(1-t)\,dt = x^2y(x)$

Step 5: Differentiate again with respect to $x$:

$y(x)(1-x) = 2xy(x) + x^2y'(x)$

Step 6: Simplify:

$y(x)(1-x-2x) = x^2y'(x)$

$y(x)(1-3x) = x^2y'(x)$

$\frac{y'(x)}{y(x)} = \frac{1-3x}{x^2}$

Step 7: Integrate:

$\ln|y| = \int\left(\frac{1}{x^2} - \frac{3}{x}\right)dx = -\frac{1}{x} - 3\ln|x| + \ln C$

Step 8: Therefore:

$y(x) = \frac{C}{x^3}e^{-1/x}$ or $y(x) = \frac{Ce^{-1/x}}{x^3}$

∴ Answer: C

Question:

Step-by-Step Solution

Master Differential Equations with Mathbee