Solve Bernoulli’s equation dxdy(x)=21x3(2x2+y(x)2)y(x), such that y(1)=1: Rewrite the equation:dxdy(x)−x5y(x)=21x3y(x)3 Divide both sides by −21y(x)3:−y(x)32dxdy(x)+y(x)22x5=−x3 Let v(x)=y(x)21, which gives dxdv(x)=−y(x)32dxdy(x):dxdv(x)+2x5v(x)=−x3 Let μ(x)=e∫2x5dx=ex6/3. Multiply both sides by μ(x):ex6/3dxdv(x)+(2ex6/3x5)v(x)=−ex6/3x3 Substitute 2ex6/3x5=dxd(ex6/3):ex6/3dxdv(x)+dxd(ex6/3)v(x)=−ex6/3x3
Apply the reverse product rule fdxdg+gdxdf=dxd(fg) to the left-hand side: dxd(ex6/3v(x))=−ex6/3x3 Integrate both sides with respect to x:∫dxd(ex6/3v(x))dx=∫−ex6/3x3dxEvaluate the integrals:ex6/3v(x)=−233x23−x6Γ(32,−3x6)+c1, where c1 is an arbitrary constant. Divide both sides by μ(x)=ex6/3:v(x)=e−x6/3(233(−x6)2/3x4Γ(32,−3x6)+c1) Solve for y(x) in v(x)=y(x)21y(x)=−233x4Γ(32,−3x6)+c1(−x6)2/3ex6/63−x6 or y(x)=233x4Γ(32,−3x6)+c1(−x6)2/3ex6/63−x6
For y(x)=−233x4Γ(32,−3x6)+(−x6)2/3c1ex6/63−x6, solve for c1 using the initial conditions: Substitute y(1)=1 into y(x)=−233x4Γ(32,−3x6)+(−x6)2/3c1ex6/63−x6:−(−1)2/3c1+233r(32,−31)3−16e=1 The equation has no solution.y(x)=−233x4r(32,−3x6)+c1(−x6)2/3ex6/63−x6 cannot satisfy the initial condition, which means no solution exists. For y(x)=233x4Γ(32,−3x6)+(−x6)2/3c1ex6/63−x6, solve for c1 using the initial conditions: Substitute y(1)=1 into y(x)=233x4Γ(32,−3x6)+(−x6)2/3c1ex6/63−x6:
(−1)2/3c1+233Γ(32,−31)3−16e=1Solve the equation:c1=61(21(123e+32/3Γ(32,−31))+23i63Γ(32,−31)) Substitute c1=61(21(123e+32/3Γ(32,−31))+23i63Γ(32,−31)) into y(x)=233x4Γ(32,−3x6)+(−x6)2/3c1ex6/63−x6y(x)=2×32/3x4Γ(32,−3x6)+(63Γ(32,−31)(3i+3)+123e)(−x6)2/323ex6/63−x6 Collecting the solutions, we have:y(x)=2×32/3x4Γ(32,−3x6)+(63Γ(32,−31)(3i+3)+123e)(−x6)2/323ex6/63−x6
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