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We can separate the variables to get

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\\[\\frac{dy}{dx}= \\simplify[std]{(1+y^2)/({a}+{b}x)} \\Rightarrow \\frac{1}{1+y^2}\\frac{dy}{dx}=\\simplify[std]{{a}+{b}x}\\]

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On integrating we get:
\\[\\arctan(y)=\\frac{1}{\\var{b}}\\ln\\left(\\left|\\var{a}+\\var{b}x\\right|\\right)+A \\Rightarrow y=\\tan\\left(\\frac{1}{\\var{b}}\\ln\\left(\\left|\\var{a}+\\var{b}x\\right|\\right)+A\\right)\\]
To fix the arbitrary constant of integration we use the condition $y(1)=\\var{u}$.

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As $\\arctan(\\var{u})=\\var{v}$ we see that $\\displaystyle{A = \\var{v}-\\frac{1}{\\var{b}}\\ln(|\\var{a+b}|)}$.

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Hence the solution is

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\\[y=\\simplify[std]{tan(({((t * (t -1)) / 2)} * (pi / 3)) + ({((t -1) * (t -2)) / 2} * (pi / 4)) + ((1 / {b}) * ln(abs(({a} + ({b} * x)) / {a + b}))))}\\]

\n ", "parts": [{"prompt": "\n

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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Input all numbers as integers or fractions.

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Separate the variables:

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Find the solution of:

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\\[\\frac{dy}{dx}=\\simplify[std]{(1+y^2)/({a}+{b}x)}\\]
which satisfies $y(1)=\\var{u}$

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Note that if $\\pi$ is in your expression you enter it as pi.

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Find the solution of $\\displaystyle \\frac{dy}{dx}=\\frac{1+y^2}{a+bx}$ which satisfies $y(1)=c$

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rebelmaths

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