Mathematics, 22.06.2019 21:00 luzcastellanos556

Find the general solution of the given differential equation. x dy dx − y = x2 sin(x) give the largest interval over which the general solution is defined. (think about the implications of any singular points. enter your answer using interval notation.


Answer from: katwilf1771

x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x

Divide both sides by x^2. In doing so, we force any possible solutions to exist on either (-\infty,0) or \boxed{(0,\infty)} (the "positive" interval in such a situation is usually taken over the "negative" one) because x cannot be 0 in order for us to do this.

\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x

Condense the left side as the derivative of a product, then integrate both sides and solve for y:

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac yx\right]=\sin x

\dfrac yx=\displaystyle\int\sin x\,\mathrm dx

\boxed{y=Cx-x\cos x}

Answer from: marykm03p3sd80

x\,\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x\implies\dfrac1x\,\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x

Note that in order to do this division, we cannot allow x=0. This means the largest interval on which a solution can exist is either (0,\infty) or (-\infty,0).

If y(x) is a solution to the ODE, then any term that vanishes as x\to\infty (or -\infty, depending on which interval above is used) is a transient term.

Solve the ODE:

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac yx\right]=\sin x\implies\dfrac yx=C-\cos x\implies y=Cx-x\cos x

As x\to\infty, \cos x will oscillate between -1 and 1, so x\cos x will oscillate between -\infty and \infty, so the limit of y(x) as x\to\infty does not exists. There are no transient terms.

Answer from: baileymtamayo

Divide both sides by x^2 - note that this means we can't have x=0:

x\dfrac{\mathrm dy}{\mathrm dx}-y=x^2\sin x

\dfrac1x\dfrac{\mathrm dy}{\mathrm dx}-\dfrac1{x^2}y=\sin x

Then the left side reduces to the derivative of a product,

\dfrac{\mathrm d}{\mathrm dx}\left[\dfrac1xy\right]=\sin x

\dfrac1xy=\displaystyle\sin x\,\mathrm dx=\cos x+C

y=x\cos x+Cx

This solution is continuous everywhere, but accounting for the singular point x=0, the largest interval over which it is defined would be (0,\infty) or (-\infty,0).

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