Answer to Question #93098 in Differential Equations for Subhasis Padhy

Question

Solve the differential equation

(z+z³) cosx dx - (z+z³) dy + (1-z²) (y-sinx) dz = 0

Answer 1

Let $u=\sin x$ . We have $\cos x dx= du$ and $(z+z^3)du-(z+z^3)dy+(1-z^2)(y-u)dz=0$ . That is $(z+z^3)(du-dy)+(1-z^2)(y-u)dz=0$ .

Next, let $u-y=v$ . Since $dv=du-dy$ , we obtain $(z+z^3)dv+(1-z^2)(-v)dz=0.$

Separate variables: $\frac{dv}{v}=\frac{1-z^2}{z+z^3}dz$ We know that $\frac{dv}{v}=d\ln |v|$ . Consider $\frac{1-z^2}{z+z^3}$ .$z+z^3=z(z^2+1)$ , so a partial fraction decomposition of $\frac{1-z^2}{z+z^3}$ is $\frac{A}{z}+\frac{Bz+C}{z^2+1}=$ $=\frac{A(z^2+1)+z(Bz+C)}{z(z^2+1)}$ and therefore $1-z^2=A(z^2+1)+z(Bz+C)=Az^2+A+Bz^2+Cz$ .

Equality of the polynomials $1-z^2$ and $Az^2+A+Bz^2+Cz=(A+B)z^2+Cz+A$ gives us a system of equations

$\begin{cases} A+B=-1\\ C=0\\ A=1 \end{cases}$ So, $A=1, B=-2, C=0$ , and $\frac{1-z^2}{z+z^3}=\frac{1}{z}-\frac{2z}{z^2+1}$ . We have $\frac{1-z^2}{z+z^3}dz=\frac{dz}{z}-\frac{2zdz}{z^2+1}=$

$=d\ln |z|-\frac{dz^2}{z^2+1}=d\ln |z|-\frac{d(z^2+1)}{z^2+1}=$

$=d\ln |z|-d\ln |z^2+1|=d\ln\bigl|\frac{z}{z^2+1}\bigr|$ .

Since $d\ln |v| = d\ln\bigl|\frac{z}{z^2+1}\bigr|$ , we have $\ln |v| = \ln\bigl|\frac{z}{z^2+1}\bigr| + D$ . Take the exponent of the equation: $|v| = e^D\bigl|\frac{z}{z^2+1}\bigr|$ . That is, $v = e^D\frac{z}{z^2+1}$ or $v = -e^D\frac{z}{z^2+1}$ . The case $v=0$ also is a solution of the equation $(z+z^3)dv+(1-z^2)(-v)dz=0.$

Union of the cases $v = e^D\frac{z}{z^2+1}$, $v = -e^D\frac{z}{z^2+1}$ and $v=0$ is the case $v=E\frac{z}{z^2+1}$ .

Remembering that $v=u-y$ and $u=\sin x$ , we have $\sin x - y = E\frac{z}{z^2+1}$

Answer: $\sin x - y = E\frac{z}{z^2+1}$