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

Question

Let $y=y(x)$ be the solution of the differential equation $({x}^{2}-x\sqrt{{x}^{2}-1})dy+(y(x-\sqrt{{x}^{2}-1})-x)dx=0,x\ge 1$. If $y(1)=1$, then the greatest integer less than $y(\sqrt{5})$ is $____$ .

Answer: 3

Accepted Answer

First write the given differential equation in the standard linear form.

$\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x-\sqrt{{x}^{2}-1}}$

Now simplify the right side by rationalising:

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

So the equation becomes:

$\frac{dy}{dx}+\frac{y}{x}=x+\sqrt{{x}^{2}-1}$

This is a linear differential equation with integrating factor (I.F.):

I.f $={e}^{\ln x}=x$

Multiply the whole equation by $x$:

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

Now integrate term by term:

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

Use the initial condition $y(1)=1$ to find $C$.

Given $y(1)=1$

Substitute $x=1$ in the expression for $yx$:

$1=\frac{1}{2}+C	o C=\frac{1}{2}$

So,

$yx=\frac{{x}^{2}}{2}+\frac{({x}^{2}-1{)}^{3/2}}{3}+\frac{1}{2}$

Divide by $x$:

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

Now put $x=\sqrt{5}$:

$[y(\sqrt{5})]=[\frac{\sqrt{5}}{2}+\frac{8}{3}+\frac{1}{2\sqrt{5}}]=3$

Solutions · 1