Question

(dy/dx) + y/sqrt{x}=x what is the integrating factor

Answer 1

Answer:

$e^{2\sqrt{x}}$

Step-by-step explanation:

$\frac{dy}{dx} + \frac{y}{\sqrt{x}} =x$

For a linear DE: $y' +P(x).y = Q(x)$

the Integrating factor is $e^{\int P(x)\,dx}$

So here the P(x) is $\frac{1}{\sqrt{x}}$

Thus the integrating factor is :

$= e^{\int P(x)\,dx} \\=e^{\int \frac{1}{\sqrt{x}}\,dx}\\=e^{\int x^{-\frac{1}{2}} \,dx }\\=e^{\frac{x^{\frac{1}{2}}}{\frac{1}{2}}}\\=e^{2\sqrt{x}}$