Question

Solve the differential equation:



x dy/dx + y = x cosx + sinx

given that y(pie/2) =1.

Answer 1

Answer:

$\text{The solution is }\bf{xy=x\:sinx}$

Step-by-step explanation:

$x\;\frac{dy}{dx}+y= x\:cosx+sinx$

$x\;dy+y\:dx= (x\:cosx+sinx)dx$....(1)

$\boxed{\begin{minipage}{5cm} Take,\:u=xy\\ \\\frac{du}{dx}=x\:\frac{dy}{dx}+y.1\\ \\ \implies\:du=x\:dy+y\:dx \end{minipage}}$

$\boxed{\begin{minipage}{5cm} Take,\:v=x\:sinx\\ \\ \frac{dv}{dx}=x\:cosx+sinx.1\\ \\ \implies\:dv=(x\:cosx+sinx)dx \end{minipage}}$

Now, (1) becomes

$du=dv$

Integrating

$\int{du}=\int{dv}$

$u=v+c$

$xy=x\:sinx+c$

$when\:x=\frac{\pi}{2},\:y=1$

$\implies\:\frac{\pi}{2}=\frac{\pi}{2}sin\frac{\pi}{2}+c$

$\implies\:\frac{\pi}{2}=\frac{\pi}{2}+c$

$\implies\:c=0$

$\therefore\text{The solution is }\boxed{\bf{xy=x\:sinx}}$

Answer 2

Solution :

The given diffentalial equation is

x dy/dx + y = x cosx + sinx

or, x dy + y dx = (x cosx + sinx) dx

or, d (xy) = d (x sinx) ...(i)

where d (xy) = x dy + y dx

and d (x sinx) = (x cosx + sinx) dx

Now, integrating from (i), we get

∫ d (xy) = ∫ d (x sinx)

or, xy = x sinx + c ...(ii)

where c is integral constant

Given that, y (π/2) = 1,

i.e., when x = π/2, y = 1

Putting x = π/2, y = 1 in (ii), we get

π/2 * 1 = π/2 * sin(π/2) + c

or, c = 0

Hence, from (ii), we get the required integral as

xy = x sinx (Ans.)