If y=y(x) is the solution of the differential equation (dy)/(dx)+4x(x^2-1)y=x+2(x^2-1)^(5)…

Question

If $y=y(x)$ is the solution of the differential equation

$\frac{dy}{dx}+\frac{4x}{({x}^{2}-1)}y=\frac{x+2}{{({x}^{2}-1)}^{\frac{5}{2}}},x>1$ such that

$y(2)=\frac{2}{9}{\log }_{e}(2+\sqrt{3})y(\sqrt{2})=\alpha {\log }_{e}(\sqrt{\alpha }+\beta )+\beta -\sqrt{\gamma },\alpha ,\beta ,\gamma \in N\alpha \beta \gamma $ :

Answer: 6

Accepted Answer

We can solve the given differential equation using an integrating factor.

The integrating factor is given by :

$\mu (x)={e}^{\int \frac{4x}{{x}^{2}-1}dx}={e}^{2\ln ({x}^{2}-1)}=({x}^{2}-1{)}^{2}$

Multiplying both sides of the differential equation by $\mu (x)$, we get :

$({x}^{2}-1{)}^{2}\frac{dy}{dx}+4xy=\frac{x+2}{{({x}^{2}-1)}^{\frac{1}{2}}}$

We can rewrite the left-hand side using the product rule:

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

Integrating both sides with respect to $x$, we get:

$({x}^{2}-1{)}^{2}y=\int \frac{x+2}{{({x}^{2}-1)}^{\frac{1}{2}}}dx=\sqrt{{x}^{2}-1}+2\ln |x+\sqrt{{x}^{2}-1}|+C$

where $C$ is the constant of integration. Using the initial condition $y(2)=\frac{2}{9}{\log }_{e}(2+\sqrt{3})$, we can solve for $C$:

At $x=2$,

$9\cdot \frac{2}{9}\ln (2+\sqrt{3})=2\ln (2+\sqrt{3})+\sqrt{3}+C$

$C=-\sqrt{3}$

At $x=\sqrt{2}$

$y(\sqrt{2})=2\ln (1+\sqrt{2})+1-\sqrt{3}$

$\beta =1,\alpha =2,\gamma =3$

$\Rightarrow \alpha \beta \gamma =6$

Solutions · 1