Question

Solve the following differential equation X(dy/dx)+2y=(x^2)logx?

Answer 1

`y = x^2/4 (logx - 1/4) + C/x^2`

Explanation:

`x(dy/dx)+2y=(x^2)logx`

Set up for integrating factor:

`dy/dx +bb(2/x)y=x logx`

The integrating factor:

`exp( int bb(2/x) dx) = exp( log x^2) = x^2`

Multiply each side by integrating factor:

`x^2dy/dx +2x y=x^3 logx`

So that:

`(x^2 \ y)^' =x^3 logx`

Integrating the RHS:

`int x^3 logx \ dx`

`= int logx \ d(x^4/4)`

By IBP:

`= x^4/4 logx - int d( logx) x^4/4`

`= x^4/4 logx - int x^3/4 dx`

`= x^4/4 logx - x^4/16 + C`

`implies x^2 \ y = x^4/4 logx - x^4/16 + C`

`y = x^2/4 (logx - 1/4) + C/x^2`

Answer 2

`y=\frac{x^2\ln x}{4}-\frac{x^2}{16}+\frac{C}{x^2}`

Explanation:

Given that

`x\frac{dy}{dx}+2y=x^2\lnx`

#\frac{dy}{dx}+ frac{2}{x}y=x\lnx#

Comparing above equation with the standard form of linear D.E. `\frac{dy}{dx}+P(x)y=Q(x)`, we get

`P(x)=2/x, \ Q(x)=x\lnx`

Integration factor (I,F.)
`I=e^{\int P(x)\ dx}=e^{\int 2/x\ dx}=e^{2\ln x}=e^{\lnx^2}=x^2`

hence, the complete solution of linear D.E. is given as follows

`y(I.F.)=\int (I.F.) Q(x)\ dx+C`

`y(x^2)=\int (x^2)x\lnx\ dx+C`

`x^2y=\int x^3\ln x\ dx+C`

`x^2y=\ln x\int x^3\ dx-\int( \frac{d}{dx}\ln x\cdot \int x^3\ dx ) dx+C`

`x^2y=\ln x\cdot \frac{x^4}{4}-\int( \frac{1}{x}\frac{x^4}{4} ) dx+C`

`x^2y=\frac{x^4\ln x}{4}-1/4\int x^3 dx+C`
`x^2y=\frac{x^4\ln x}{4}-1/4\frac{x^4}{4}+C`
`x^2y=\frac{x^4\ln x}{4}-\frac{x^4}{16}+C`

`y=\frac{x^2\ln x}{4}-\frac{x^2}{16}+\frac{C}{x^2}`

Answer 3

`y = 1/4x^2lnx - 1/16x^2 + C/x^2`

Explanation:

First, we are going to divide the entire equation by `x` to put it into the form `dy/dx + color(red)(P(x))y = color(blue)(Q(x))`.

`dy/dx + color(red)(2/x)y = color(blue)(xlnx)`

Now, we need to find the special integrating factor. For a differential equation in this form, the special integrating factor is given by

`mu=e^{intcolor(red)(P(x))dx}`

`mu=e^{int color(red)(2/x) dx}`

`mu=e^{2lnx}`

`mu=(e^lnx)^2`

`color(green)(mu=x^2)`

So we may now multiply the original equation by `color(green)(mu)`.

`color(green)(x^2)(dy/dx + 2/x y) = color(green)(x^2)(xlnx)`

`x^2dy/dx + 2xy = x^3lnx`

If this technique is new to you, the reason that we've multiplied by the special integrating factor is to form the result of a derivative of a product on the left-hand side.

`d/dx(x^2y) = x^3lnx`

Now, we are almost done. To find `y`, we should first integrate both sides. Recall that integrating a derivative results in the original function.

`int (d/dx(x^2y))dx = color(orange)(int (x^3lnx) dx)`

I've just written an antiderivative for the orange integral below, but I'll show you the work for it at the very end.

`x^2y = 1/4x^4lnx - 1/16x^4 + C`

And isolating `y`:

`y = 1/4x^2lnx - 1/16x^2 + C/x^2`

---

`color(orange)(int (x^3lnx) dx)`

Integrate by substitution. Let `color(red)(x = e^t), color(red)(lnx = t), color(red)(dx = e^t dt )`.

`int ((color(red)(e^t))^3color(red)t) color(red)(e^tdt)`

`= int(t * e^(3t) * e^t)du`

`= int(te^(4t))dt`

Perform integration by parts.

`u = t`, `du = dt`
`dv=e^(4t)dt, v = 1/4e^(4t)`

`uv - int vdu`

`= 1/4te^(4t) - int (1/4e^(4t))dt`

`= 1/4te^(4t) - 1/16e^(4t) + C`

`= 1/4t(e^t)^4 - 1/16(e^t)^4 + C`

Undo substitution `color(red)(x = e^t), color(red)(lnx = t)`.

`= 1/4 * lnx * (x)^4 - 1/16(x)^4 + C`

`= 1/4x^4lnx - 1/16x^4 + C`