Question

Find the general solution of the given differential equation.

Answer 1

Answer:

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

There are no transient terms.

Step-by-step explanation:

Given: $x\,\frac{dy}{dx} -y=x^2\,sin\,x$

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

Solution:

For an equation of the form $\frac{dy}{dx}+yp(x)=q(x)$,

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

The given equation $x\,\frac{dy}{dx} -y=x^2\,sin\,x$ can be written as $\frac{dy}{dx}-\frac{y}{x}=x\,sin\,x$

Here,

$p(x)=\frac{-1}{x}\,,\,q(x)=x\,sin\,x$

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

So,

the solution is $\frac{y}{x}=\int \frac{1}{x}x\,sin\,x\,dx$

$\frac{y}{x} =\int\,sin\,x\,dx\\\\\frac{y}{x} =-cos\,x+C\\y=-x\,cos\,x+Cx$

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.