Question

Solve 𝑥2𝑦``−3𝑥𝑦`+5𝑦=𝑥2sin(𝑙𝑛𝑥) ;𝑦(1)=1,𝑦`(1)=0

Answer 1

Solution;

This is a Cauchy Euler differential equation;

Substitute $x=e^t$ hence t=ln(x)

$\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=e^{-t}\frac{dy}{dt}$

$\frac{d^2y}{dx^2}=\frac{d}{dx}(e^{-t}\frac{dy}{dt})=-e^{-2t}\frac{dy}{dt}+e^{-2t}\frac{d^2y}{dt^2}$

Substitute in the given equation;

$e^{2t}(-e^{-2t}\frac{dy}{dt}+e^{-2t}\frac{d^2y}{dt^2})-3e^t(e^{-t}\frac{dy}{dt})+5y=e^{2t}sin(t)$

Simplify;

$\frac{d^2y}{dt}-4\frac{dy}{dt}+5y=e^{2t}sin(t)$

Above equation is homogeneous;

The characteristic equation ;

$r^2+4r+5=(r-2)^2+1$

The roots are;

r=2+i and r=2-i

The homogeneous solution;

$y_h=e^{2t}(C_1cos(t)+C_2sin(t))$

The particular solution;

$y_p=t.(e^{2t}Acos t+e^{2t}Bsint)$

By distribution;

$y_p=te^{2t}(Acost+Bsint)$

Differentiating;

$y'_p=(2te^{2t}+e^{2t})(Acost+Bsint)+te^{2t}(-Asint+Bcost)$

Differentiate ;$y''_p=(4te^{2t}+4e^{2t})(Acost+Bsint)+(4te^{2t}+2e^{2t})(-Asint+Bcost)+te^{2t}(-Acost-Bsint)$

Substitute into the transformed equation and simplify;

$e^{2t}(-2Asint +2Bcost)=e^{2t}sint$

Equate the coefficients;

-2A=1;A=-0.5

B=0

Thefore;

$y_p=-\frac12te^{2t}cost$

Since ;

$y=y_h+y_p$

The general solution of the equation is ;

$y=e^{2t}(C_1cost+C_2sint)-\frac12te^{2t}cost$

Write in terms of x by replacing x=e^t;

$y=x^2(C_1cos(lnx)+C_2sin(lnx))-\frac12x^2ln(x)cos(lnx)$

Apply the initial conditions to find the constants;

y(1)=1

1=$C_1$

$y=x^2cos(lnx)+C_2x^2sin(lnx)-\frac12x^2lnxcos(lnx)$

$y''=C_2(x^2sin(lnx)-\frac{x^2lnxcos(lnx)}{2})+x^2cos(lnx)$

Use the condition;

y'(1)=0

0=C₂(0+0)+0

C₂=0

Hence;$y=x^2cos(lnx)-\frac12x^2ln(x)cos(lnx)$