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Find the solution of a first order separable differential equation of the form $(a+x)y'=b+y$.
"}, "statement": "Find the solution of the differential equation
\n\\[(\\var{a1}+x)y'=\\var{b1}+y,\\]
\nsatisfying $y(\\var{c1})=\\var{d1}$.
", "advice": "The differential equation is separable, meaning it can be rearranged to give all the terms in $x$ on one side, and all the terms in $y$ on the other.
\nSo we can write
\n\\[\\int{\\!\\frac{1}{\\var{b1}+y}\\,\\mathrm{d}y} = \\int{\\!\\frac{1}{\\var{a1}+x}\\,\\mathrm{d}x},\\]
\nthen
\n\\[\\ln\\lvert\\var{b1}+y\\rvert=\\ln\\lvert\\var{a1}+x\\rvert+c,\\]
\nExponentiating each side gives:
\n\\[e^{\\ln\\lvert\\var{b1}+y\\rvert}=e^{\\ln\\lvert\\var{a1}+x\\rvert+c},\\]
\nso
\n\\[y=\\simplify{A({a1}+x)-{b1}},\\]
\nwhich is the general solution of the equation.
\nNow,
\n\\[\\var{d1}=y(\\var{c1})=\\simplify[std]{A({a1}+{c1})-{b1}},\\]
\nso
\n\\[A=\\simplify[std]{({d1}+{b1})/({a1}+{c1})}=\\simplify{{d1+b1}/{a1+c1}},\\]
\nand then the full solution is
\n\\[y=\\simplify[std]{{d1+b1}/{a1+c1}({a1}+x)-{b1}}=\\simplify{{(d1*a1-b1*c1)}/{a1+c1}+{(d1+b1)}*x/{a1+c1}}.\\]
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