Let y=y(x) be the solution of the differential equation (dy)/(dx)=(1+x+x^2)(1-y+y^2),y(0)=…

Question

Let $y=y(x)$ be the solution of the differential equation $\frac{dy}{dx}=(1+x+{x}^{2})(1-y+{y}^{2}),y(0)=\frac{1}{2}$. Then $(2y(1)-1)$ is equal to

Accepted Answer

Given

$\frac{dy}{dx}=(1+x+{x}^{2})(1-y+{y}^{2}),\, y(0)=\frac{1}{2}$

We will solve it by the method of separation of variables.

Since

$\frac{dy}{1-y+{y}^{2}}=(1+x+{x}^{2})\, dx$

Now simplify the denominator on the left:

$1-y+{y}^{2}={y}^{2}-y+1={(y-\frac{1}{2})}^{2}+\frac{3}{4}$

So,

$\int \frac{dy}{{(y-\frac{1}{2})}^{2}+{(\frac{\sqrt{3}}{2})}^{2}}=\int (1+x+{x}^{2})\, dx$

Using the standard result

$\int \frac{du}{{u}^{2}+{a}^{2}}=\frac{1}{a}{	an }^{-1}(\frac{u}{a})$

we get

$\frac{2}{\sqrt{3}}{	an }^{-1}(\frac{2y-1}{\sqrt{3}})=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}+C$

Now use the initial condition $y(0)=\frac{1}{2}$.

Then

$2y-1=0$

so

$\frac{2}{\sqrt{3}}{	an }^{-1}(0)=C\Rightarrow C=0$

Hence,

$\frac{2}{\sqrt{3}}{	an }^{-1}(\frac{2y-1}{\sqrt{3}})=x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}$

At $x=1$,

$x+\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}=1+\frac{1}{2}+\frac{1}{3}=\frac{11}{6}$

Therefore,

$\frac{2}{\sqrt{3}}{	an }^{-1}(\frac{2y(1)-1}{\sqrt{3}})=\frac{11}{6}$

So

${	an }^{-1}(\frac{2y(1)-1}{\sqrt{3}})=\frac{11\sqrt{3}}{12}$

Taking tangent on both sides,

$\frac{2y(1)-1}{\sqrt{3}}=	an (\frac{11\sqrt{3}}{12})$

Hence,

$2y(1)-1=\sqrt{3}	an (\frac{11\sqrt{3}}{12})$

So the correct answer is

$\sqrt{3}	an (\frac{11\sqrt{3}}{12})$

Therefore, the correct option is .

Solutions · 1