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

Question

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

Evelyn Agard · Accepted Answer

#2# You got #y = 1+C_0/sqrt(x^2+1)# Now, the condition #y(0) = 4/3 = 1+C_0/sqrt(0^2+1)# gives #C_0 = 1/3# or #y = 1+1/3 1/sqrt(x^2+1)# and finally #y(sqrt8)+8/9 = 1+1/3 1/sqrt(8+1) + 8/9=2#

Christopher Allers · Answer

undefined 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 $y(\sqrt{8}) + \frac{8}{9}$:

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

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

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 #(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