Answer to Question #199577 in Differential Equations for rajkumar

Question

solve the following differential equations -

sin^-1(dy/dx)= x+y

(1+y)²dx=(tan^-1y-x)dy

(D-1)²(D²+1)² Y= sin² (x/2) + e^x + x

2x²y(d²y/dx²) + 4y²= x² ( dy/dx)² +2xy (dy/dx)

Answer 1

1)

$y'=sin(x+y)$

$x+y=v$

$v'-1=y'$

$v'-1=sinv$

$\frac{dv}{1+sinv}=dx$

$\frac{dv}{1+sinv}=\int \frac{sec^2(v/2)}{(tan(v/2)+1)^2}dv=|u=tan(v/2)+1,dv=\frac{2}{sec^2(v/2)}du|=$

$=\int 2 du/u^2=-2/u=-\frac{2}{tan(v/2)+1}$

$-\frac{2}{tan((x+y)/2)+1}=x+c$

2)

$R=(y+1)^2,S=x-tan^{-1}y$

find integrating factor $\mu(y)$ such that

$\mu R+y'\mu S=0$ is exact

then:

$\frac{\partial}{\partial y}(\mu R)=\frac{\partial}{\partial x}(\mu S)$

$d\mu /dy(y+1)^2+2(y+1)\mu =\mu$

$\frac{d\mu /dy}{\mu}=\frac{-2y-1}{(y+1)^2}$

$ln\mu =-2ln(y+1)-\frac{1}{y+1}$

integrating factor:

$\mu(y)=\frac{e^{-1/(y+1)}}{(y+1)^2}$

then:

$e^{-1/(y+1)}+\frac{(e^{-1/(y+1)}(x-tan^{-1}y))y'}{(y+1)^2}=0$

$P(x,y)=e^{-1/(y+1)},Q(x,y)=\frac{e^{-1/(y+1)}(x-tan^{-1}y)}{(y+1)^2}$

$f(x,y)=\int P(x,y)dx=\int e^{-1/(y+1)}dx=xe^{-1/(y+1)}+g(y)$

$\frac{\partial f(x,y)}{\partial y}=\frac{xe^{-1/(y+1)}}{(y+1)^2}+\frac{dg(y)}{dy}$

$\frac{\partial f(x,y)}{\partial y}=Q(x,y)$

$\frac{xe^{-1/(y+1)}}{(y+1)^2}+\frac{dg(y)}{dy}=\frac{e^{-1/(y+1)}(x-tan^{-1}y)}{(y+1)^2}$

$\frac{dg(y)}{dy}=-\frac{e^{-1/(y+1)}tan^{-1}y}{(y+1)^2}$

$g(y)=\int -\frac{e^{-1/(y+1)}tan^{-1}y}{(y+1)^2}dy=-e^{-1/(y+1)}-\frac{1}{2}ie^{-1/2+i/2}Ei(-\frac{1}{y+1}+\frac{1-i}{2})+$

$+\frac{1}{2}ie^{-1/2-i/2}Ei(-\frac{1}{y+1}+\frac{1+i}{2})--e^{-1/(y+1)}(tan^{-1}y-1)$

$f(x,y)=xe^{-1/(y+1)}-e^{-1/(y+1)}-\frac{1}{2}ie^{-1/2+i/2}Ei(-\frac{1}{y+1}+\frac{1-i}{2})+$

$+\frac{1}{2}ie^{-1/2-i/2}Ei(-\frac{1}{y+1}+\frac{1+i}{2})--e^{-1/(y+1)}(tan^{-1}y-1)$

solution:

$f(x,y)=c$

$xe^{-1/(y+1)}-e^{-1/(y+1)}-\frac{1}{2}ie^{-1/2+i/2}Ei(-\frac{1}{y+1}+\frac{1-i}{2})+$

$+\frac{1}{2}ie^{-1/2-i/2}Ei(-\frac{1}{y+1}+\frac{1+i}{2})--e^{-1/(y+1)}(tan^{-1}y-1)=c$

where Ei is exponential integral:

$Ei(x)=\int_{-\infin}^x e^t dt/t$

3)

characteristic equation:

$(k-1)^2(k^2+1)^2=0$

$k_{1,2}=1$

$k_{3,4,5,6}=\pm i$

$y_c=c_1e^x+c_2xe^x+c_3cosx+c_4sinx+c_5xcosx+c_6xsinx$

for particular solution:

$y_{p1}=\frac{e^x}{(1+1)^2}\frac{x^2}{2!}=x^2e^x/4$

$sin^2 (x/2)=\frac{1-cosx}{2}$

$y_{p2}=-\frac{1}{2(-2-2D)(D^2+1)}\frac{x}{2}\int cosxdx=\frac{1}{8(1+D)(D^2+1)}xsinx$

$\frac{1}{(D^2+1)}xsinx=\frac{1}{(D^2+1)}\cdot I.P.(xe^{ix})= I.P.e^{ix}\frac{1}{2Di}(1-\frac{D}{2i}+...)x=$

$I.P.e^{ix}\frac{1}{2Di}(x-\frac{1}{2i})= I.P.e^{ix}\frac{1}{2i}(x^2/2-\frac{x}{2i})=$

$= I.P.(x^2/2-\frac{x}{2i})(cosx+isinx)=xcosx/2+x^2sinx/2$

where I.P. is Imaginary part

$y_{p2}=\frac{1}{8(1+D)}(xcosx/2+x^2sinx/2)$

$\frac{1}{(1+D)}xcosx=R.P.e^{ix}\frac{1}{(1+D+i)}x=R.P.e^{ix}(\frac{1-i}{2}+iD/2-...)x=$

$=R.P.((x-ix+i))(cosx+isinx)/2=(xcosx+(x-1)sinx)/2$

where R.P. is real part

$\frac{1}{(1+D)}x^2sinx=I.P.e^{ix}(\frac{1-i}{2}+iD/2-(1+i)D^2/4+...)x^2=$

$=I.P.(cosx+isinx)((x^2-ix^2)/2+ix-(1+i)/2)=$

$=-x^2cosx/2+xcosx-cosx/2+x^2sinx/2-sinx/2$

$y_{p2}=\frac{1}{32}((x-2)sinx -(x^2+1)cosx+3xcosx+x^2sinx)$

$y_{p3}=\frac{1}{(D-1)^2(D^2+1)^2}(x+1/2)=(1+2D+...)^2(1-D^2+...)^2(x+1/2)=$

$=(1+2D+...)(x+5/2)=x+9/2$

$y=c_1e^x+c_2xe^x+c_3cosx+c_4sinx+c_5xcosx+c_6xsinx+x^2e^x/4+$

$+\frac{1}{32}((x-2)sinx -(x^2+1)cosx+3xcosx+x^2sinx)+x+9/2$

4)

$2x^2yy''+4y^2=x^2(y')^2+2xyy'$

$y=x^2z^2$

then:

$y=x^2z^2$

$y'=2xz^2+2x^2zz'$

$y''=2z^2+4xzz'+4xzz'+2x^2((z')^2+zz'')$

$2x^2x^2z^2(2z^2+4xzz'+4xzz'+2x^2((z')^2+zz''))+4x^4z^4=$

$=x^2(2xz^2+2x^2zz')^2+2xx^2z^2(2xz^2+2x^2zz')$

$x^2z^2(z^2+4xzz'+x^2((z')^2+zz''))+x^2z^4=$

$=x^2z^4+2x^3z^3z'+x^4z^2(z')^2+xz^2(xz^2+x^2zz')$

$z^2+4xzz'+x^2((z')^2+zz'')+z^2=z^2+2xzz'+x^2(z')^2+z^2+xzz'$

$xzz'+x^2zz''=0$

$xz(z'+xz'')=0$

$z'+xz''=0$

$z'=u$

$u+xu'=0$

$du/u=-dx/x$

$lnu=-lnx+lnc_1$

$u=c_1/x$

$dz=c_1dx/x$

$z=c_1lnx+c_2$

$y(x)=x^2(c_1lnx+c_2)^2$