Question #93098
Solve the differential equation

(z+z³) cosx dx - (z+z³) dy + (1-z²) (y-sinx) dz = 0
1
Expert's answer
2019-08-26T09:02:48-0400

Let u=sinxu=\sin x . We have cosxdx=du\cos x dx= du and (z+z3)du(z+z3)dy+(1z2)(yu)dz=0(z+z^3)du-(z+z^3)dy+(1-z^2)(y-u)dz=0 . That is (z+z3)(dudy)+(1z2)(yu)dz=0(z+z^3)(du-dy)+(1-z^2)(y-u)dz=0 .

Next, let uy=vu-y=v . Since dv=dudydv=du-dy , we obtain (z+z3)dv+(1z2)(v)dz=0.(z+z^3)dv+(1-z^2)(-v)dz=0.

Separate variables: dvv=1z2z+z3dz\frac{dv}{v}=\frac{1-z^2}{z+z^3}dz We know that dvv=dlnv\frac{dv}{v}=d\ln |v| . Consider 1z2z+z3\frac{1-z^2}{z+z^3} .z+z3=z(z2+1)z+z^3=z(z^2+1) , so a partial fraction decomposition of 1z2z+z3\frac{1-z^2}{z+z^3} is Az+Bz+Cz2+1=\frac{A}{z}+\frac{Bz+C}{z^2+1}= =A(z2+1)+z(Bz+C)z(z2+1)=\frac{A(z^2+1)+z(Bz+C)}{z(z^2+1)} and therefore 1z2=A(z2+1)+z(Bz+C)=Az2+A+Bz2+Cz1-z^2=A(z^2+1)+z(Bz+C)=Az^2+A+Bz^2+Cz .

Equality of the polynomials 1z21-z^2 and Az2+A+Bz2+Cz=(A+B)z2+Cz+AAz^2+A+Bz^2+Cz=(A+B)z^2+Cz+A gives us a system of equations

{A+B=1C=0A=1\begin{cases} A+B=-1\\ C=0\\ A=1 \end{cases} So, A=1,B=2,C=0A=1, B=-2, C=0 , and 1z2z+z3=1z2zz2+1\frac{1-z^2}{z+z^3}=\frac{1}{z}-\frac{2z}{z^2+1} . We have 1z2z+z3dz=dzz2zdzz2+1=\frac{1-z^2}{z+z^3}dz=\frac{dz}{z}-\frac{2zdz}{z^2+1}=

=dlnzdz2z2+1=dlnzd(z2+1)z2+1==d\ln |z|-\frac{dz^2}{z^2+1}=d\ln |z|-\frac{d(z^2+1)}{z^2+1}=

=dlnzdlnz2+1=dlnzz2+1=d\ln |z|-d\ln |z^2+1|=d\ln\bigl|\frac{z}{z^2+1}\bigr| .

Since dlnv=dlnzz2+1d\ln |v| = d\ln\bigl|\frac{z}{z^2+1}\bigr| , we have lnv=lnzz2+1+D\ln |v| = \ln\bigl|\frac{z}{z^2+1}\bigr| + D . Take the exponent of the equation: v=eDzz2+1|v| = e^D\bigl|\frac{z}{z^2+1}\bigr| . That is, v=eDzz2+1v = e^D\frac{z}{z^2+1} or v=eDzz2+1v = -e^D\frac{z}{z^2+1} . The case v=0v=0 also is a solution of the equation (z+z3)dv+(1z2)(v)dz=0.(z+z^3)dv+(1-z^2)(-v)dz=0.

Union of the cases v=eDzz2+1v = e^D\frac{z}{z^2+1}, v=eDzz2+1v = -e^D\frac{z}{z^2+1} and v=0v=0 is the case v=Ezz2+1v=E\frac{z}{z^2+1} .

Remembering that v=uyv=u-y and u=sinxu=\sin x , we have sinxy=Ezz2+1\sin x - y = E\frac{z}{z^2+1}


Answer: sinxy=Ezz2+1\sin x - y = E\frac{z}{z^2+1}


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