(4x^3+3x^2+3y-1)dx+(3x+2y+1)dy=0

Lösung

(4 x^{3} + 3 x^{2} + 3 y - 1) d x + (3 x + 2 y + 1) d y = 0

y = \frac{- 1 - 3 x + - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}, y = \frac{- 1 - 3 x - - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}

Schritte zur Lösung

(4 x^{3} + 3 x^{2} + 3 y - 1) d x + (3 x + 2 y + 1) d y = 0

Solve exact ODE:

y = \frac{- 1 - 3 x + - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}, y = \frac{- 1 - 3 x - - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}

y = \frac{- 1 - 3 x + - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}, y = \frac{- 1 - 3 x - - 4 x ^{4} - 4 x ^{3} + 9 x ^{2} + 10 x + c _{1} + 1}{2}

x \to \infty lim (\frac{5 e ^{- 3 x} + 2}{e ^{- 5 x} + 4})

\int (5 x + 3) 3 5 x^{2} + 6 x d x

y^{^{''}} + 7 y^{^{'}} + 10 y = 4 t e^{4 t}

n = 1 \sum \infty (cos (1.4))^{n}

\int_{0}^{t} t^{2} e^{- 5 t} d t