2y = x - 4 4x + 3y = 5

Mathematics

1 answer:

MakcuM [25]3 years ago

8 0

Answer:

the answer is, {x,y} ={2,1}

Step-by-step explanation:

1) 4x - 3y = 5 2) x + 2y = 4 3) 4•(-2y+4) - 3y = 5 4) - 11y = -11 5) 11y = 11 y = 1 6) x = -2y+4 y = 1 7) x = -2(1)+4 = 2 = {x,y} = {2,1} !!!

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

Tpy6a [65]

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)

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