de-CH
utf-8
math math-format
Separation of Variables: Solution Function
dgl-06-033
multiple
72
randRangeNonZero(-6, 6) randRange(2, 6)

Determine the general solution of the ODE y'(x) = \dfrac{x}{\sqrt{B + x^2}}(y(x) + A)^2 and the constant \color{red}C in case of the initial value y(0) = 0 .

x y(x) = - \frac{1}{\sqrt{B + x^2} +C} - A
c \color{red}C = -sqrt(B) - 1/A

We look for the general solution using the separation of variables.

We rewrite the ODE as: y'(x) = \dfrac{dy}{dx} = \dfrac{x}{\sqrt{B + x^2}}(y(x) + A)^2 \implies \dfrac{1}{(y(x) + A)^2} \ dy = \dfrac{x}{\sqrt{B + x^2}} \ dx.

Now we look for antiderivatives \displaystyle \int\dfrac{1}{(y(x) + A)^2} \ dy = - \dfrac{1}{(y(x) + A)} = \int \dfrac{x}{\sqrt{B + x^2}} \ dx = \sqrt{B + x^2} + C, by using substution on the right-hand side.

We want to solve the equation for y(x), i.e. try to isolate it:

Taking the reciprocal and subtracting terms yields \displaystyle y(x) =- \dfrac{1}{\sqrt{B + x^2} +{\color{red}C}} - A.

For the constant C we set x=0 and use the inital value y(0) = 0:

\displaystyle y(0) = 0 = - \dfrac{1}{\sqrt{B} +C} - A \implies {\color{red} C = - \sqrt{B} - \frac{1}{A}} = -sqrt(B)-1/A (at least approximately . . . .)