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blxxmgrxcie
30.07.2019 •
Mathematics
Find the general solution to 1/x dy/dx - 2y/x^2 = x cos x, y(pi) = pi^2
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Ответ:
Step-by-step explanation:
Consider linear differential equation![\frac{\mathrm{d} y}{\mathrm{d} x}+yp(x)=q(x)](/tpl/images/0151/9076/fedf5.png)
It's solution is of form
where I.F is integrating factor given by
.
Given:![\frac{1}{x}\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x^2}=x\cos x](/tpl/images/0151/9076/75a84.png)
We can write this equation as![\frac{\mathrm{d} y}{\mathrm{d} x}-\frac{2y}{x}=x^2\cos x](/tpl/images/0151/9076/84b77.png)
On comparing this equation with
, we get ![p(x)=\frac{-2}{x}\,\,,\,\,q(x)=x^2\cos x](/tpl/images/0151/9076/a10b3.png)
I.F =
{ formula used:
}
we get solution as follows:
{ formula used:
}
Applying condition:![y(\pi)=\pi^2](/tpl/images/0151/9076/a5525.png)
So, we get solution as :
Ответ:
Step-by-step explanation:
55?