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Question: In Problems 1-30, solve the equation.

x2-3y2dx+2xydy=0

Short Answer

Expert verified

The solution of the given equation isy2=Cx3+x2 and x = 0.

Step by step solution

01

Given information and simplification

Given that, x2-3y2dx+2xydy=01

Evaluate the equation (1).

x2-3y2dx+2xydy=02xydy=-x2-3y2dx

dydx=-x2-3y22xy2

Since the given equation is homogeneous. Then, put y = tx and dydx=xdtdx+t.

Substitute the values in equation (2).

role="math" localid="1663948749390" dydx=-x2-3y22xyxdtdx+t=-1-3t22txdtdx=-1-3t22t-txdtdx=-1+3t2-2t22t=-1+t22t2tt2-1dt=dxx

Let us assume p = t2 - 1. Differentiate with respect to x.

dp = 2t dt .

Substitute the values in equation (1)

1pdp=dxx

02

Evaluation method

Now integrate the equation (2) on both sides.

1pdp=dxxlnp=lnx+Cp=Cx

Substitute the value of p.

t2-1=Cx

Substitute the value of t.

y2-x2=Cx3y2=Cx3+x2

So, the solution is y2=Cx3+x2

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