de-CH
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Separation of Variables: Solution Value
dgl-06-01
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) randRange(1, 4) randRangeExclude(2,6,[5]) M-1 "\\frac{" + N + "}{" + M + "}" "\\frac{" + 1 + "}{" + M + "}" M pow(M,M) (pow(A*V*X1*X1/2 + A * W*X1,IK) - B*PK)/(A*PK) fractionReduce(pow(A*V*X1*X1/2 +A* W*X1,IK) - B*PK,A*PK)

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

Determine the value f(X1).

x f(X1) = Snum

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 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 first determine the constant C=0, applying x = 0 and the given initial condition f(0) = Y0.

Finally, we solve the equation for y(x):

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

We then substitute X1 into this function and obtain as the solution f(X1) = S.