Question

1)(2-у) dy = xdx, х=2, у=2; 2) dy/dx = 2х+1, х=1,у=2.

Answer 1

Ответ:

$4y-y^{2}-x^{2}=C_{1}, C_{1}=2C, C-const; 4y-y^{2}-x^{2}=0;$

$y-x^{2}-x=C, C-const; y-x^{2}-x=0;$

Объяснение:

$1) (2-y)dy=xdx;$

$\int\ {(2-y)} \, dy = \int\ {x} \, dx ;$

$\int\ {2} \, dy - \int\ {y} \, dy = \frac{x^{1+1}}{1+1}+C, C-const;$

$2y-\frac{y^{1+1}}{1+1}=\frac{x^{2}}{2}+C, C-const;$

$2y-\frac{1}{2}y^{2}-\frac{1}{2}x^{2}=C, C-const;$

$4y-y^{2}-x^{2}=C_{1}, C_{1}=2C, C-const;$

$x=2, y=2;$

$C_{1}=4*2-2^{2}-2^{2}=8-4-4=0;$

$4y-y^{2}-x^{2}=0;$

$2) \frac{dy}{dx}=2x+1;$

$dy=(2x+1)dx;$

$\int\ {1} \, dy = \int\ {(2x+1)} \, dx ;$

$y = \int\ {2x} \, dx + \int\ {1} \, dx ;$

$y=2\int\ {x} \, dx +x+C, C-const;$

$y=2*\frac{x^{1+1}}{1+1}+x+C, C-const;$

$y=2*\frac{x^{2}}{2}+x+C, C-const;$

$y=x^{2}+x+C, C-const;$

$y-x^{2}-x=C, C-const;$

$x=1, y=2;$

$C=2-1^{2}-1=2-1-1=0;$

$y-x^{2}-x=0;$

Answer 2

It looks like you've provided two different mathematical equations involving derivatives. Let's address each of them:

Equation 1:

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(2 - y) dy = x dx

Given that x = 2 and y = 2, let's find the solution to this equation.

Integrate both sides with respect to their respective variables:

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∫(2 - y) dy = ∫x dx

Integrating the left side:

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2y - 0.5y^2 = 0.5x^2 + C

Substitute the values x = 2 and y = 2:

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2(2) - 0.5(2)^2 = 0.5(2)^2 + C
4 - 2 = 2 + C
C = 0

So, the equation becomes:

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2y - 0.5y^2 = 0.5x^2
2y - 0.5y^2 = 0.5(2)^2
2y - 0.5y^2 = 2
2y - 0.5y^2 - 2 = 0

This is a quadratic equation in terms of y. You can solve it using the quadratic formula.

Equation 2:

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dy/dx = 2x + 1

Given that x = 1 and y = 2, you're looking for the value of the derivative dy/dx at x = 1, y = 2.

Substitute the values x = 1 and y = 2 into the equation:

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dy/dx = 2(1) + 1
dy/dx = 3

So, the value of the derivative dy/dx at x = 1, y = 2 is 3.

If you have any further questions or if I misunderstood your inquiry, please feel free to ask!