# If #(1+x^2)dy/dx=x(1-y) and y(0)=4/3 #,then the value of #y(sqrt(8))+8/9# is ?

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If #(1+x^2)dy/dx=x(1-y)
and
y(0)=4/3 # ,then the value of #y(sqrt(8))+8/9# is

After solving the differential equation I get

#ysqrt(1+x^2)=sqrt(1+x^2)+C#

what is to be done after this

If

After solving the differential equation I get

what is to be done after this

You got

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To solve the given differential equation ((1+x^2)\frac{dy}{dx} = x(1-y)) with the initial condition (y(0) = \frac{4}{3}), we first separate the variables and then integrate both sides.

[ (1+x^2)dy = x(1-y)dx ]

[ \int (1+x^2)dy = \int x(1-y)dx ]

Integrating both sides gives:

[ y + \frac{x^3}{3} = \frac{x^2}{2} - \frac{x^2y}{2} + C ]

Now, we use the initial condition (y(0) = \frac{4}{3}) to find the constant (C):

[ \frac{4}{3} + 0 = 0 - 0 + C ] [ C = \frac{4}{3} ]

Substitute (C = \frac{4}{3}) back into the equation:

[ y + \frac{x^3}{3} = \frac{x^2}{2} - \frac{x^2y}{2} + \frac{4}{3} ]

Now, we can solve for (y) at (x = \sqrt{8}):

[ y + \frac{8\sqrt{8}}{3} = \frac{16}{2} - \frac{16y}{2} + \frac{4}{3} ]

Simplify and solve for (y):

[ y + \frac{8\sqrt{8}}{3} = 8 - 8y + \frac{4}{3} ] [ y + 8y = 8 - \frac{4}{3} - \frac{8\sqrt{8}}{3} ] [ 9y = \frac{20}{3} - \frac{8\sqrt{8}}{3} ] [ y = \frac{20 - 8\sqrt{8}}{27} ]

Finally, evaluate (y(\sqrt{8}) + \frac{8}{9}):

[ y(\sqrt{8}) + \frac{8}{9} = \frac{20 - 8\sqrt{8}}{27} + \frac{8}{9} ] [ = \frac{60 - 24\sqrt{8} + 72}{243} ] [ = \frac{132 - 24\sqrt{8}}{243} ]

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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