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In problems 33-40, Solve the equation given in Problem 1.

2txdx+(t2-x2)dt=0

Short Answer

Expert verified

x2+t2-Ct=0andt0

Step by step solution

01

Solving the given equation

The given equation is,

2txdx+t2-x2dt=0

Simplify the above equation,

dxdt=-t2-x22txdxdt=x2t-t2x

Let,

localid="1663928794004" v=xtx=vtdxdt=v+tdvdt

Substitute xt=vthe dxdt=v+tdvdtand in the equation (1),

v+tdvdt=v2-12v

Simplify the above equation,

tdvdt=v2-12v-vtdvdt=v2-1-2v22vtdvdt=-1-v22vtdvdt=-1+v22v

Cross multiplication on both sides in the above equation,

tdvdt=-1+v22v2v1+v2dv=-dtt

02

Integrating

Taking integration both sides in the above equation

2v1+v2dv=-dtt

Solve the above equation,

ln1+v2=-lnt+lnCln1+v2+lnt=lnC1+v2t=C

03

Finding the solution

Substitute the v=xt in the above equation,

1+xt2t=C

Simplify the above equation,

t2+x2tt2=Cx2+t2=Ctx2+t2-Ct=0

Also note that is t=0 a solution

Hence the solutionx2+t2-Ct=0 is andt0

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