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$\\displaystyle{x\\frac{dy}{dx}+\\var{a}y=\\simplify[std]{{b}x^{n}}}$ can be solved by the integrating factor method.
\nFirst we divide both sides by $x$ to obtain $\\displaystyle{\\frac{dy}{dx}+\\frac{\\var{a}}{x}y=\\simplify[std]{{b}x^{n-1}}}$
\nThis is now in the standard form $\\displaystyle{\\frac{dy}{dx}+a(x)y=b(x)}$ which we can solve by multiplying through by the integrating factor $e^{\\int a(x)dx}$
\n\nIn our example, $a(x) = \\frac{\\var{a}}{x}$ so $e^{\\int a(x)dx}=e^{\\int{\\frac{\\var{a}}{x}dx}}=e^{\\var{a}\\ln(x)}=x^\\var{a}$
\n\nHence multiplying both sides of $\\displaystyle{\\frac{dy}{dx}+\\frac{\\var{a}}{x}y=\\simplify[std]{{b}x^{n-1}}}$ by $x^\\var{a}$ gives
\n$\\displaystyle{x^\\var{a}\\frac{dy}{dx}+\\var{a}x^{\\var{a-1}}y=\\simplify[std]{{b}x^{n-1+a}}}$
\nWe note that we can rewrite the left hand side as $\\frac{d(x^{\\var{a}}y)}{dx}$ (product rule) and hence:
\n\\[\\frac{d(x^{\\var{a}}y)}{dx}=\\simplify[std]{{b}x^{a+n-1}}\\]
We can integrate both sides to get:
\\[x^{\\var{a}}y=\\simplify[std]{{b}/{a+n}x^{a+n}+A}\\]
to determine $A$ we use the condition $\\displaystyle{y(1)=\\simplify[std]{{b*(c+1)}/{a+n}}}$ and we see that:
\\[\\simplify[std]{{b*(c+1)}/{a+n}}=\\simplify[std]{{b}/{a+n}+A}\\]
\\[A = \\simplify[std]{{b*(c+1)}/{a+n}} - \\simplify[std]{{b}/{a+n} = {b*c}/{a+n}}\\]
and so the solution is:
\\[x^{\\var{a}}y=\\simplify[std]{{b}/{a+n}x^{a+n}+{b*c}/{a+n}} \\Rightarrow y=\\simplify[std]{{b}/{a+n}*(x^{n}+{c}*x^{-a})}\\]
Input all numbers as integers or fractions.
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\n$y=\\;\\;$[[0]]
\nInput all numbers as integers or fractions – not as decimals.
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\nFind the solution of:
\\[x\\frac{dy}{dx}+\\var{a}y=\\simplify[std]{{b}x^{n}}\\]
which satisfies $\\displaystyle{y(1)=\\simplify[std]{{b*(c+1)}/{a+n}}}$
", "type": "question", "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}]}], "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}, {"name": "Maria Aneiros", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3388/"}]}