Differential equations for high numbers, differential equations for high numbers

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The force of this question should be the resultant force, if it is not the resultant force, it cannot be calculated, and the question should be handed in clearly. Here's how to solve it:

Let the force on the object be f, the velocity of the object is v, and the acceleration of the object is a=dv dt, which is determined by Newton's second law, f=ma=m*dv dt

From known conditions: f=kt v, where k is a proportionality constant, not difficult to obtain, k=1. Therefore.

t/v=m*dv/dt=dv/dt

vdv=tdt

Integral, v 2=t 2+c, where c is the integral constant. When t=1, v=8m s, so c=63.

v^2=t^2+63

When v=12m s, t=9s can be obtained.

Let f=kt v, when t=1s, v=8m s, f=(1 8)nSubstitution:

1 8=k 8, so k=1;So f=t v.

Let the acceleration be am s; Then f=ma=a, (m=1kg), that is, there is t v=a=dv dt;

Separation of variables yields VDV=TDT, and integralization yields (1 2)V = (1 2)T +(1 2)C;

i.e. v = t +c; Substituting the initial conditions t=1 and v=8 into the solution yields c=63

Therefore, the velocity v = t +63

Therefore, when v=12m s, there is t =12 -63=144-63=81; ∴t=9s.

That is, when t = 9 seconds, the velocity v=12m s

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Can be turned into:

yy'=e^(y^2)*e^(3x)

Separate variables. y*e (-y 2)dy=e (3x)dx.

y*e^(-y^2)dy=∫e^(3x)dx∫-1/2d[e^(-y^2)]=∫1/3d(e^(3x))-1/2*e^(-y^2)=1/3*e^(3x)+c