y^'=4+e^{y-4x+1}

解答

y^{^{'}} = 4 + e^{y - 4 x + 1}

y = - ln (- x - c_{1}) + 4 x - 1

求解步骤

y^{'} (x) = 4 + e^{y - 4 x + 1}

使用替代法求解:

y = - ln (- x - c_{1}) + 4 x - 1

y = - ln (- x - c_{1}) + 4 x - 1

in v erse l a pl a ce \frac{400}{s ^{2} + 12 s + 400}

t an g e n t f (x) = x^{2} + 4, (- 3, 13)

\int \frac{cos ^{3} ( 3 t )}{sin ^{5} ( 3 t )} d t

y^{^{'}} + \frac{2 y}{t - 2} = \frac{sin ( t )}{t - 2}

\frac{\partial}{\partial x} (3 x y + 4 x z + 5 yz - 3)