Question

Please solve the linear differential equations
1. dx/y+z = dy/-(x+z) = dz/x-y

2.(x^2D^2-3xD+5)y =x^2(logx)

Answer 1

$$Solution$$

1. dx/y+z = dy/-(x+z) = dz/x-y

We solve it y Lagrange's multipliers method

$\frac{(ldx+mdy+ndz)}{(lP+mQ+nR)}=0\\ lP+mQ+nR=0\\$

So,

$ldx+mdy+ndz=0$

Here $l,m,n$ are multipliers

$\frac{dx}{y+z}=\frac{dy}{-x-z}=\frac{dz}{x-y},$ given.

So,

$1dx+1dy+1dz=0$

Since

$1(y+z)+1(-x-z)+1(x-y)=0,$

$1,1,1$ are multipliers

And

$x(y+z)+y(-x-z)+z(x-y)=0,$

So $x,y,z$ are multipliers

$\therefore 1dx+1dy+1dz=0$

After integration

$x+y+z=C_1---------(i)$

Also $xdx+ydy+zdz=0$

After integration

$x^2+y^2+z^2=C_2---------(ii)$

$(i)$ and $(ii)$ are the required solutions

2.(x^2D^2-3xD+5)y =x^2(logx)

Let $x=e^z$ and $ln x=z$

$\implies \frac{dx}{dz}=e^z=x\\ \implies \frac{dz}{dx}=\frac{1}{x}$

Now,

$\frac{dy}{dx}=\frac{dy}{dz} \cdot \frac{dz}{dx} = \frac{dy}{dz} \cdot \frac{1}{x}\\ x \frac{dy}{dx} = \frac{dy}{dz}\\ if\ \frac{d}{dz} \equiv \theta, then\\ xDy= \theta y$

Similarly,

$x^2D^2y=\theta(\theta - 1) y$

Now,

$(\theta ^2 - \theta - 3\theta+5)y=ze^{2z}\\ \implies (\theta ^2 - 4\theta+5)y=ze^{2z}$

This is a non homogeneous differential equation of the second order.

The remaining solution should be elementary.

The characteristic equation $r^2-4r+5$ has the solutions $r=2-i, 2+i$ . While the particular solution (by the method of undetermined coefficients) comes out to be $Y_p=e^{2z}$

Then the general solution (in terms of $z$ ) is given by,

$$y=y_c+y_p=C_1e^{z(2+i)}+C_2e^{z(2-i)}+e^{2z}$$

for arbitrary constants $C_1$ and $C_2.$

$$y=C_1x^{(2+i)}+C_2x^{(2-i)}+x^{2}$$