Answer to Question #268193 in Differential Equations for Code_X_

Question

 Using Coefficient Linear in Two Variables, determine the general solution of (𝒙 − 𝟑𝒚 − 𝟑)𝒅𝒙 + 𝟑(𝒙 + 𝟑𝒚 − 𝟑)𝒅𝒚 = 0

Answer 1

Solution;

$(x-3y-3)dx+(3x+9y-9)dy=0....(1)$

From (1) we have two linear equations;

$a)x-3y-3$

$b)3x+9y-9$

Solve for the intersection (h,k) using (a) and (b);

$a)x-3y-3=0$

From which ;

$x=3+3y$

$b)3x+9y-9=0$

Substitute x;

$3(3y+3)+9y-9=0$

$9y+9+9y-9=0$

$18y=0$

y=0

Hence;

$x=3(0)+3=3$

Now,(h,k)=(3,0)

Take;

$x=u+h=u+3$

$y=v+k=v$

And;

$dx=du$

$dy=dv$

Substitute into (1);

$[(u+3)-3(v)-3]du+[3(u+3)+9(v)-9]dv=0$

Simplify into;

$(u-3v)du+(3u-9v)dv=0$ ...(2)

The above is an homogeneous equation ,we pick M(u,v). Let;

u=mv

$du=mdv+vdm$

Substitute into (2);

$(mv-3v)(mdv+vdm)+(3mvdv-9vdv)=0$

Simplifies to;

$m^2vdv+mv^2dm-3v^2dm+9vdv)=0$

$v(m^2-9)dv+v^2(m-3)dm=0$

Resolve as;

$\frac{dv}{v}+\frac{m-3}{m^2-9}dm=0$

Simplifies to;

$\frac{dv}{v}+\frac{dm}{m+3}=0$

Integrate;

$\int\frac1vdv+\int\frac{1}{m+3}dm=0$

$ln(v)+ln(m+3)=lnc$

$lnv(m+3)=lnc$

Multiply by exponential;

$v(m+3)=c$

But ;

$u=mv\implies m=\frac uv$

Therefore;

$v(\frac uv+3)=c$

But;

$x=u+3\implies u=x-3$

And ;

$y=v$

Now;

$u+3v=c$

Becomes;

$(x-3)+3y=0$