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How can I check that $\;y=\dfrac{\sin x}{x}\;$ is a solution of $\;xy'+y=\cos x\;$?

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    Ah yea..Of course. Sometimes the solution is so simple that I tend to overlook it. Thanks.2012-12-17

5 Answers 5

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Using the quotient rule for differentiation, we have $xy'+y=x\left(\frac{x\cos x-\sin x}{x^2}\right)+y=\cos x-\frac{\sin x}{x}+\frac{\sin x}{x}=\cos x.$

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    Very nice, thank you.2012-12-17
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Compute $y'$ using $y=\dfrac{\sin x}{x},\tag{1}$

Substitute your result, $y'$, into the equation $xy'+y=\cos x,\tag{2}$

Evaluate, and check to confirm that the left-hand side of $(2)$ equals the right-hand side of $(2)$.

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One can always check whether a function "works" by substituting. In this case, we can do better. For note that $xy'+y=(xy)'$. So our equation can be rewritten as $(xy)'=\cos x.$ Integrate. We get $xy'=\sin x +C$ and therefore the general solution is $\frac{\sin x +C}{x}.$

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In case you wanted to know how to arrive at the solution. Since this is a linear equation of order one:

$y'x+y=\cos x\Longrightarrow y'+\frac{1}{x}y=\frac{\cos x}{x}$

We now put

$\mu(x):=e^{\int\frac{dx}{x}}=e^{\log x}=x\Longrightarrow$

The general solution is

$y=\frac{1}{x}\left(\int\frac{\cos x}{x}\cdot e^{\int\mu(x)dx}dx\right)+C=\frac{1}{x}\int\cos xdx+C=\frac{\sin x}{x}+C\,\,,\,C=\,\text{a constant}$

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Just differentiate $\Rightarrow xy=\sin(x)$

applying uv rule on L.H.S and $\sin$ derivative on R.H.S

$\Rightarrow xy'+ y=\cos( x)$