(8-r^2)dr=r^3sin(θ)dθ

Lösung

(8 - r^{2}) d r = r^{3} sin (θ) d θ

θ = arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn, θ = - arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn

Schritte zur Lösung

(8 - r^{2}) d r = r^{3} sin (θ) d θ

Solve separable ODE:

θ = arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn, θ = - arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn

θ = arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn, θ = - arccos (- \frac{- r ^{2} ln ( r ) - 4 + c _{1} r ^{2}}{r ^{2}}) + 2 πn

\int_{0}^{1} \frac{x - 6}{x ^{2} - 6 x + 8} d x

3 x \frac{d y}{d x} = x^{2} - 4

x \to 4 lim (2 x^{2} - 2 x)

\int_{- \infty}^{6} r e^{\frac{r}{2}} d r

\frac{\partial}{\partial x} (4 y^{3})