Question

Find a general solution to the following first-order differential equation for y(x): dy/dx - y/x = x^2

Answer 1

The general solution of the following first-order differential equation for $y(x) : \frac{dy}{dx}-\frac{y}{x}=x^{2}$  is  $x^{3}+cx$.

We have to find the general solution of the given first order differential equation.

Let y = u*v, where u and v are functions of x.

Hence,

dy/dx = u*dv/dx + v*du/dx

Putting this in the given equation, we get,

$u\frac{dv}{dx}+v\frac{du}{dx}-\frac{uv}{x}=x^{2}\\\\ u\frac{dv}{dx}+v(\frac{du}{dx}-\frac{u}{x} )=x^{2}$

Putting the v term as 0, we get

du/dx - u/x = 0

du/dx = u/x

du/u = dx/x

Integrating both sides, we get

$\int {\frac{1}{u} } } \, du=\int {\frac{1}{x} } \, dx$

ln(u) = ln(x) + p

Let p = ln(k)

Hence,

ln(u) = ln(x) + ln(k)

ln(u) = ln(kx)

u = kx

Putting u = kx in the differential equation with v term 0 , we get

$(kx)\frac{dv}{dx} =x^{2}\\\\(k)dv=\frac{x^{2}dx}{x} \\\\(k)dv=(x)dx\\\\ \int {k} \, dv=\int {x}dx\\\\kv = x^{2}+c\\\\v=\frac{x^{2}+c}{k}$

Putting v in y = uv, we get

$y=kx*(\frac{x^{2}+c}{k} )\\\\y=x(x^{2}+c)\\\\y=x^{3}+cx$

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