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Verify that when the linear differential equation Pxy-Qxdx+dy=0is multiplied by x=ePxdx, the result is exact.

Short Answer

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Step by step solution

01

General form of special integrating factors

Theorem 3:

IfMy-NxN is continuous and depends only on x, thenx=expMy-NxNdx

is an integrating factor for the equation.

IfNx-MyM is continuous and depends only on y, theny=expNx-MyMdy

is an integrating factor for the equation.

02

Evaluation method

Given that, Pxy-Qxdx+dy=01

Multiply byx=ePxdxon both sides. We get,

PxyePxdx-QxePxdxdx+ePxdxdy=02PxyePxdx-QxePxdxdx+ePxdxdy=02

Let M=PxyePxdx-QxePxdx,N=ePxdx.

Then,

My=PxePxdxNx=PxePxdx

Therefore, .

My=Nx

So, the result for this problem is exact.

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