If y = y(x) is the solution of the differential equation, x[dy dx] + 2y = x2, satisfying y…

Question

If y = y(x) is the solution of the differential equation,

x$$dy \over dx$$ + 2y = x2, satisfying y(1) = 1, then y($$1\over2$$) is equal
to :

Accepted Answer

Given,

$$x{{dy} \over {dx}} + 2y = {x^2}$$

$$ \Rightarrow $$  $${{dy} \over {dx}} + \left( {{2 \over x}} \right)y = x$$

This is a linear differential equation.

$$ \therefore $$  I.F $$ = {e^{\int {{2 \over x}dx} }}$$

$$ = {e^{2\ln x}}$$

$= {x^2}$

$$ \therefore $$  Solution is,

$$y \cdot {x^2} = \int {x \cdot {x^2}dx} $$

$$ \Rightarrow $$  $$y{x^2} = {{{x^4}} \over 4} + C$$

given  $$y\left( 1 \right) = 1$$

$$ \therefore $$  $$1.1 = {4 \over 4} + C$$

$$ \Rightarrow $$  $$C = {3 \over 4}$$

$$ \therefore $$  Equation is

$$y{x^2} = {{{x^4}} \over 4} + {3 \over 4}$$

$$ \therefore $$  $$y\left( {{1 \over 2}} \right)$$   means   $$x = {1 \over 2}$$

$$ \therefore $$  $$y \cdot {\left( {{1 \over 2}} \right)^2} = {1 \over 4} \times {\left( {{1 \over 2}} \right)^4} + {3 \over 4}$$

$$ \Rightarrow $$  $${y \over 4} = {1 \over {64}} + {3 \over 4}$$

$$ \Rightarrow $$   $${y \over 4} = {{1 + 48} \over {64}}$$

$$ \Rightarrow $$  y = $${{49} \over {16}}$$

Solutions · 1