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Paladinen [302]
3 years ago
15

Solve the following differential equation: (2x+5y)dx+(5x−4y)dy=0 *Hint: they are exact C=.

Mathematics
1 answer:
Tpy6a [65]3 years ago
5 0

Answer with Step-by-step explanation:

The given differential equation is

(2x+5y)dx+(5x-4y)dy=0

Now the above differential equation can be re-written as

P(x,y)dx+Q(x,y)dy=0

Checking for exactness we should have

\frac{\partial P}{\partial y}=\frac{\partial Q}{\partial x}

\frac{\partial P}{\partial y}=\frac{\partial (2x+5y)}{\partial y}=5

\frac{\partial Q}{\partial x}=\frac{\partial (5x-4y)}{\partial x}=5

As we see that the 2 values are equal thus we conclude that the given differential equation is exact

The solution of exact differential equation is given by

u(x,y)=\int P(x,y)dx+\phi(y)\\\\u(x,y)=\int (2x+5y)dx+\phi (y)\\\\u(x,y)=x^2+5xy+\phi (y)

The value of \phi (y) can be obtained by differentiating u(x,y) partially with respect to 'y' and equating the result with P(x,y)

\frac{\partial u}{\partial y}=\frac{\partial (x^2+5xy+\phi (y)))}{\partial y}=Q(x,y))\\\\5y+\phi '(y)=(5x-4y)\\\\\phi '(y)=5x-9y\\\\\int\phi '(y)\partial y=\int (5x-9y)\partial y\\\\\phi (y)=5xy-\frac{9y^2}{2}\\\\\therefore u(x,y)=x^2+10xy-\frac{9y^2}{2}+c

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