(d^2x)/(dt^2)5(-qdx/dt-kxp)

 Esercizio

$\frac{d^2x}{dt^2}\left(5\cdot\left(-q\frac{dx}{dt}-kx+p\right)\right)$

 

Applicare la formula: $x\left(a+b\right)$$=xa+xb$, dove $a=-q\left(\frac{dx}{dt}\right)$, $b=-kx+p$, $x=5$ e $a+b=-q\left(\frac{dx}{dt}\right)-kx+p$

$\frac{d^2x}{dt^2}\left(\frac{-5qdx}{dt}+5\left(-kx+p\right)\right)$

 

Applicare la formula: $x\left(a+b\right)$$=xa+xb$, dove $a=-kx$, $b=p$, $x=5$ e $a+b=-kx+p$

$\frac{d^2x}{dt^2}\left(\frac{-5qdx}{dt}-5kx+5p\right)$

 

Applicare la formula: $x\left(a+b\right)$$=xa+xb$, dove $a=\frac{-5qdx}{dt}$, $b=-5kx+5p$, $x=\frac{d^2x}{dt^2}$ e $a+b=\frac{-5qdx}{dt}-5kx+5p$

$\frac{d^2x}{dt^2}\frac{-5qdx}{dt}+\frac{d^2x\left(-5kx+5p\right)}{dt^2}$

 

Applicare la formula: $x\left(a+b\right)$$=xa+xb$, dove $a=-5kx$, $b=5p$, $x=d^2$ e $a+b=-5kx+5p$

$\frac{d^2x}{dt^2}\frac{-5qdx}{dt}+\frac{\left(-5d^2kx+5d^2p\right)x}{dt^2}$

 

Applicare la formula: $x\left(a+b\right)$$=xa+xb$, dove $a=-5d^2kx$, $b=5d^2p$ e $a+b=-5d^2kx+5d^2p$

$\frac{d^2x}{dt^2}\frac{-5qdx}{dt}+\frac{-5x^2d^2k+5xd^2p}{dt^2}$

$\frac{d^2x}{dt^2}\frac{-5qdx}{dt}+\frac{-5x^2d^2k+5xd^2p}{dt^2}$

 Come posso risolvere questo problema?

Non riuscite a trovare un metodo? Segnalatecelo, così potremo aggiungerlo.