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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.

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First we divide both sides by $x$ to obtain $\\displaystyle{\\frac{dy}{dx}+\\frac{\\var{a}}{x}y=\\simplify[std]{{b}x^{n-1}}}$

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This 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 $e^{\\int a(x)dx}$

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In our example, $a(x) = \\frac{\\var{a}}{x}$ so $e^{\\int a(x)dx}=e^{\\int{\\frac{\\var{a}}{x}dx}}=e^{\\var{a}\\ln(x)}=x^\\var{a}$

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Hence multiplying both sides of $\\displaystyle{\\frac{dy}{dx}+\\frac{\\var{a}}{x}y=\\simplify[std]{{b}x^{n-1}}}$ by $x^\\var{a}$ gives

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$\\displaystyle{x^\\var{a}\\frac{dy}{dx}+\\var{a}x^{\\var{a-1}}y=\\simplify[std]{{b}x^{n-1+a}}}$

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We note that we can rewrite the left hand side as $\\frac{d(x^{\\var{a}}y)}{dx}$ (product rule) and hence:

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\\[\\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:

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\\[\\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})}\\]

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Input all numbers as integers or fractions.

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Solution is:

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$y=\\;\\;$[[0]]

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Input all numbers as integers or fractions – not as decimals.

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Find the solution of $\\displaystyle x\\frac{dy}{dx}+ay=bx^n,\\;\\;y(1)=c$

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rebelmaths

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Integrating Factor Method:

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Find the solution of:
\\[x\\frac{dy}{dx}+\\var{a}y=\\simplify[std]{{b}x^{n}}\\]

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which satisfies $\\displaystyle{y(1)=\\simplify[std]{{b*(c+1)}/{a+n}}}$

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