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Using the IF to find the General Solution.

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Integrating Factor Method:

Find the General Solution of:
\\[\\simplify{{A}x^{{m}}}\\simplify{(d y)/(d x)+{B}}x^\\simplify{{n}+{m}}y=\\simplify{{D}x}^\\simplify{{n}+{m}}\\]

Enter your answer as a formula: use the syntax c*e^(n*x^p) for $ce^{nx^p}$, not forgetting the * for multiplying arbitrary constants. Use lowercase $c$ for any constant of integration.

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\$\\simplify{{A}x^{{m}}}\\simplify{(d y)/(d x)+{B}}x^\\simplify{{n}+{m}}y=\\simplify{{D}x}^\\simplify{{n}+{m}}\$ can be solved by the integrating factor method.

Firstly divide both sides by \$\\simplify{{A}x^{{m}}}\$ to obtain \$\\simplify{(d y)/(d x)+{B}/{A}}\\simplify{x^{n}}y=\\simplify{{D}/{A}*x^{n}}\$

This is now in the standard form $\\displaystyle{\\frac{dy}{dx}+f(x)y=r(x)}$ which can be solved using the integrating factor $I=e^{\\int f(x)dx}$

In this example, $f(x) =\\simplify{{B}/{A} x^{n}}$ so $e^{\\int f(x)dx}=e^{\\int \\simplify{{B}/{A} x^{n} d x}}=e^{\\simplify{{B}/{(n+1)*A} x^{n+1}}}$

The integrating factor formula \$ y = \\frac{1}{I}\\int r(x) I dx\$ can be used with \$ r(x) = \\simplify{{D}/{A}x^{n}} \$ to obtain:

\\[\\displaystyle y =e^{\\simplify{-{B}/{(n+1)*A} x^{n+1}}} \\int \\simplify{{D}/{A}x^{{n}}}e^{\\simplify{{B}/{({n+1})*A} x^{n+1}}} dx\\]

This can be solved using the substitution: \$ u= \\simplify{{B}/{(n+1)*A} x^{n+1}}\$ so \$ \\frac{du}{dx} = \\simplify{{B}/{A} x^{n}}\$ and therefore \$\\simplify{x^{n}} dx = \\simplify{{A}/{B}}du\$, which gives:

\\[\\displaystyle y = e^{-u} \\int \\simplify{{D}/{B}e^u} du = e^{-u} \\left[\\simplify{{D}/{B} e^u} + c\\right] = \\simplify{{D}/{B}} + ce^{-u}\\]

Substituting back \$ u= \\simplify{{B}/{(n+1)*A} x^{n+1}}\$ results in a general solution: \$ y = \\simplify{{D}/{B}} + ce^{\\simplify{-{B}/{(n+1)*A} x^{n+1}}} \$

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Solution is:

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

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

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Input all numbers as integers or fractions.

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