X^{2}[(y'−x^{2})+3xy=cosx, (x>0)

The given differential equation is

x ² (y' - x ²) + 3xy = cos(x)

Expanding and rearranging terms, we get

x ² y' + 3xy = cos(x) + x ⁴

Multiply both sides by x, which is motivated by the fact that (x ³)' = 3x ².

x ³ y' + 3x ²y = x cos(x) + x ⁵

The left side is the derivative of a product:

(x ³y)' = x cos(x) + x ⁵

Integrate both sides with respect to x :

∫ (x ³y)' dx = ∫ (x cos(x) + x ⁵) dx

x ³y = cos(x) + x sin(x) + 1/6 x ⁶ + C

Solve for y. Since x > 0, we can safely divide both sides by x ³.

y = cos(x)/x ³ + sin(x)/x ² + 1/6 x ³ + C/x³

seems like d

step-by-step explanation: