y^'+(ty)/(t^2)= 4/(t^2)

Lösung

y^{^{'}} + \frac{t y}{t ^{2}} = \frac{4}{t ^{2}}

y = \frac{4 ln ( t )}{t} + \frac{c _{1}}{t}

Schritte zur Lösung

y^{'} (t) + \frac{t y}{t ^{2}} = \frac{4}{t ^{2}}

Solve linear ODE:

y = \frac{4 ln ( t )}{t} + \frac{c _{1}}{t}

y = \frac{4 ln ( t )}{t} + \frac{c _{1}}{t}

\frac{d y}{d x} = \frac{x}{7 cos ( y )}

\frac{d}{d x} (\frac{( x + 3 ) ^{2}}{( x + 2 ) ^{5} ( x + 4 ) ^{7}})

\int co tan^{2} (x) d x

\frac{\partial}{\partial x} (\frac{1}{2 - x + y})

t an g e n t 2 x + x^{2} - 5, at x = 1