1/4 y^{''}+4t=2tan(4t)-1/3 e^t

解答

\frac{1}{4} y^{^{''}} + 4 t = 2 tan (4 t) - \frac{1}{3} e^{t}

y = - \frac{1}{4} (i Li_{2} (- e^{2 \cdot 4 i t}) + 16 i t^{2} - 8 t ln (1 + (- 1)^{\frac{8 t}{π}}) + 8 t ln (cos (4 t))) - \frac{4}{3} e^{t} - \frac{8 t ^{3}}{3} + c_{1} t + c_{2}

求解步骤

\frac{1}{4} y^{''} (t) + 4 t = 2 tan (4 t) - \frac{1}{3} e^{t}

整理

y^{''} (t) : y^{''} (t) = 8 tan (4 t) - \frac{4}{3} e^{t} - 16 t

解

y^{''} (t) = 8 tan (4 t) - \frac{4}{3} e^{t} - 16 t : y^{'} (t) = - 2 ln (cos (4 t)) - \frac{4}{3} e^{t} - 8 t^{2} + c_{1}

解

y^{'} (t) = - 2 ln (cos (4 t)) - \frac{4}{3} e^{t} - 8 t^{2} + c_{1} : y = - \frac{1}{4} (i Li_{2} (- e^{2 \cdot 4 i t}) + 16 i t^{2} - 8 t ln (1 + (- 1)^{\frac{8 t}{π}}) + 8 t ln (cos (4 t))) - \frac{4}{3} e^{t} - \frac{8 t ^{3}}{3} + c_{1} t + c_{2}

y = - \frac{1}{4} (i Li_{2} (- e^{2 \cdot 4 i t}) + 16 i t^{2} - 8 t ln (1 + (- 1)^{\frac{8 t}{π}}) + 8 t ln (cos (4 t))) - \frac{4}{3} e^{t} - \frac{8 t ^{3}}{3} + c_{1} t + c_{2}

x \to 0 lim (\frac{sec ( x ) - 1}{x})

\frac{d}{d x} (\frac{1}{4 x - 1})

\int \frac{sec ( 2 y ) tan ( 2 y )}{4 + sec ( 2 y )} d y

d er i v a t i v e - 2 x^{4} + 7 x e^{3 x - 2}

\int_{1}^{\infty} ln (x) d x