Let y = y ( x ) , y 0 be a solution curve of the differential equation ( 1 + x 2 ) ? d y = y ( x - y ) ? d x . If ( 0 ) = 1 and ( 2 2 ) = , then

Question

Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x$ . If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then

Smart Achievers · Accepted Answer

(1)

$Given, (1 + x^{2}) d y = y (x - y) d x$

$Where, y (0) = 1, y (2 \sqrt{2}) = β$

$d y = (\frac{y x - y^{2}}{1 + x^{2}}) d x$

$\frac{d y}{d x} + y (\frac{- x}{1 + x^{2}}) = (\frac{- 1}{1 + x^{2}}) y^{2}$

$Divide by y^{2} both sides and proceed.$

Solution

Q.
Let $y = y (x), y > 0$ be a solution curve of the differential equation $(1 + x^{2}) d y = y (x - y) d x$ . If $y (0) = 1$ and $y (2 \sqrt{2}) = β$ , then

1 $e^{3 β^{- 1}} = e (3 + 2 \sqrt{2})$

2 $e^{β^{- 1}} = e^{- 2} (5 + \sqrt{2})$

3 $e^{β^{- 1}} = e^{- 2} (3 + 2 \sqrt{2})$

4 $e^{3 β^{- 1}} = e (5 + \sqrt{2})$

Ans.
(1)

$Given, (1 + x^{2}) d y = y (x - y) d x$

$Where, y (0) = 1, y (2 \sqrt{2}) = β$

$d y = (\frac{y x - y^{2}}{1 + x^{2}}) d x$

$\frac{d y}{d x} + y (\frac{- x}{1 + x^{2}}) = (\frac{- 1}{1 + x^{2}}) y^{2}$

$Divide by y^{2} both sides and proceed.$

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Solution

Q. Let y=y(x), y>0 be a solution curve of the differential equation (1+x2) dy=y(x-y) dx. If y(0)=1 and y(22)=β, then

1 e3β-1=e(3+22)

2 eβ-1=e-2(5+2)

3 eβ-1=e-2(3+22)

4 e3β-1=e(5+2)

Ans. (1) Given, (1+x2) dy=y(x-y) dx Where, y(0)=1, y(22)=β dy=(yx-y21+x2)dx dydx+y(-x1+x2)=(-11+x2)y2 Divide by y2 both sides and proceed.