Question

If y=y(x) is the solution of differential equation xdy/dx + 2y = x^2 satisfying y(1) = 1 then 16y(1/2) is equal to ?

Answer 1

Answer:

friend plz attach a figure as without which we would not be able to solve it,...

Answer 2

$\frac {49}{16}$ is the Answer

Step-by-step explanation:

Given,

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

which can also be written as  

$\frac {dy}{dx} = y\frac {2}{x} = x$ (Which is Linear Differential Equation of First Order)

Now, lets find integrating factor,

which will be off the form,

$\frac {dy}{dx} + py = Q$

$e^{\int pdx} = e^{2\int\frac {1}{x}dx }$

= $e^{\frac {2}{x}} = e^{log^2} = x^2$ the integrating factor!

Now, using this, $yx^2$,

= $y.x^2 = \int x.x^2.dx + C$

= $y.x^2 = \frac {x^4}{4}.dx + C$

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

For y(1) = 1,  

$1 = \frac {1}{4} + C$

C = $1 - \frac {1}{4}$

$C = \frac {3}{4}$

Equation becomes, y = $\frac {x^2}{4} + \frac {3}{4x^2}$

For y = $\frac {1}{2}$

we have, $y(\frac {1}{2}) = \frac {1}{4} \times \frac {1}{4} + \frac {3}{4} \times 4$

= $\frac {1}{16} + 3$

= $\frac {49}{16}$