Find the general solution of the given differential equation. x dy/ dx − y = x^2 sin(x) Also determine whether there are any transient terms in the general solution. (Enter the …

Question

asked Oct 18, 2021 34.8k views

1 Answer

Lfaraone · Answer 1 · 2021-10-24T09:58:42+0000

Answer:

$y=-x\,cos\,x+Cx$

There are no transient terms.

Explanation:

Given:
$x\,(dy)/(dx) -y=x^2\,sin\,x$

To find: general solution of the differential equation and the transient terms in the general solution.

Solution:

For an equation of the form
(dy)/(dx)+yp(x)=q(x) ,

solution is given by
$ye^{\int {p(x)} \, dx }$ = ∫ q(x)
$e^{\int {p(x)} \, dx }$ dx

The given equation
$x\,(dy)/(dx) -y=x^2\,sin\,x$ can be written as
$(dy)/(dx)-(y)/(x)=x\,sin\,x$

Here,

$p(x)=(-1)/(x)\,,\,q(x)=x\,sin\,x$

$e^{\int{p(x)} \, dx } =e^{\int{(-1)/(x) } \, dx } =e^(-ln(x)) =e^{ln(x^(-1) )}=x^(-1)=(1)/(x)$

So,

the solution is
$(y)/(x)=\int (1)/(x)x\,sin\,x\,dx$

$(y)/(x) =\int\,sin\,x\,dx\\\\(y)/(x) =-cos\,x+C\\y=-x\,cos\,x+Cx$

Here, C is a constant.

Transient term is a term such that it tends to 0 as x → ∞

Here, there does not exist any term that tends to 0 as x → ∞

So, there are no transient terms.