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Separation of Variables: Solution Function
dgl-06-02
multiple
17280
0 randRangeExclude(-4, 4, [-1,0,1] ) randRangeExclude(-4, 4, [-1,0,1] ) randRangeExclude(-4, 4, [-1,0,1] ) randRangeExclude(-4, 4, [-1,0,1, A, -A] ) fractionReduce(-B, A) randRangeExclude(2,6,[5]) M-1 "\\frac{" + N + "}{" + M + "}" "\\frac{" + 1 + "}{" + M + "}" M pow(M,M)

Determine the solution of the IVP y'(x) = (V x + W) \left(A y(x) + B\right)^{\frac{N}{M}} with f(0) = Y0.

x f(x) = 1/A * (fractionReduce(A*V,2*M) * x^2 + fractionReduce(W*A,M) x )^{M} - fractionReduce(B,A)

At first, we determine the solution f with separation of variables.

We rewrite the ODE as y'(x) = \dfrac{dy}{dx} = (V x + W) \left(A y(x) + B\right)^{ K}\implies \left(A y(x) + B\right)^{-K} \ dy = (V x + W) \ dx.

Now we look for antiderivatives \displaystyle \int \left(A y(x) + B\right)^{-K} \ dy = \int (V x + W) \ dx = fractionReduce(V,2)x^2 + W x + C.

For the left-hand side \displaystyle \int \left(A y(x) + B\right)^{-K} \ dy use e.g. substitution to get as primitive \displaystyle fractionReduce(M, A) (A y(x) + B)^{KK}.

After equating \displaystyle fractionReduce(M, A) (A y(x) + B)^{KK} = fractionReduce(V,2)x^2 + W x + C we get the constant C=0, setting x = 0 and using the initial value f(0) = Y0.

To get the term for y(x) we try to isolate it:

\displaystyle fractionReduce(M, A) (A y(x) + B)^{KK} = fractionReduce(V,2)x^2 + W x \implies (A y(x) + B)^{KK} = fractionReduce(A*V,2*M)x^2 + fractionReduce(W*A,M) x \implies y(x) = fractionReduce(1,A)\left(fractionReduce(A*V,2*M)x^2 + fractionReduce(W*A,M) x \right)^{M} - fractionReduce(B,A).