Question #109878
Solve the following differential equations
i) y sin 2x dx =(1+ y² + cos² x) dy

ii) (xy² - x²) dx + ( 3x²y² + x²y - 2x³ + y²) dy =0
1
Expert's answer
2020-04-16T17:25:55-0400

¡) Ans:

The given differential equation is

ysin2xdx=(1+y2+cos2x)dyy sin2x dx=(1+y^2+cos^2x)dy

    ysin2xdx(1+y2+cos2x)dy=0\implies y sin2x dx -(1+y^2+cos^2x)dy=0

Which is of the form Mdx+Ndy=0Mdx +Ndy =0 .

Where M=ysin2x and N=(1+y2+cos2x)M=y sin2x \ \text{and} \ N=-(1+y^2+cos2x)


My=ysin2xy=sin2x\frac{\partial{M}}{\partial{y}}=\frac{\partial{ysin2x}}{\partial{y}}=sin2xNx=(1+y2+cos2x)x=(2cosx×sinx)\frac{\partial{N}}{\partial{x}}=\frac{\partial{-(1+y^2+cos^2x)}}{\partial{x}}=-(2cosx×-sinx)

=2sinxcos=sin2x=2sinxcos =sin2x

As My=Nx\frac{\partial{M}}{\partial{y}}=\frac{\partial{N}}{\partial{x}} ,Hence the given differential equation is an exact differential equation.

\therefore \ The solution of the given differential equation is

y=const.Mdx+(only those therms of N which do not contains x)dy\int_{y=const.}Mdx+\int \text{(only those therms of N which do not contains x)}dy

=y=const.ysin2x+(1+y2)dy=\int_{y=const.}ysin2x+\int{-(1+y^2)}dy


=ysin2xdx(1+y2)dy=y\int sin2x dx-\int(1+y^2)dy

=ycos2x2yy33+c=-\frac{ycos2x}{2}-y-\frac{y^3}{3}+c

Where "c" is the integration constant.



¡¡)ans :

The given differential equation is

(xy2x2)dx+(3x2y2+x2y2x3+y2)dy=0(xy^2-x^2)dx+(3x^2y^2+x^2y-2x^3+y^2)dy=0

Which is of the form Mdx+Ndy=0Mdx+Ndy=0

Here M=(xy2x2) and N=(3x2y2+x2y2x3+y)M=(xy^2-x^2) \ \text{and} \ N=(3x^2y^2+x^2y-2x^3+y)

My=2xy and Nx=6xy2+2xy6x2\therefore \frac{\partial{M}}{\partial{y}}=2xy \ \text{and} \ \frac{\partial{N}}{\partial{x}}=6xy^2+2xy-6x^2

As MyNx\text{As} \ \frac{\partial{M}}{\partial{y}}\neq \frac{\partial{N}}{\partial{x}}

Therefore ,the given differential equation is not exact.

But 1M(NxMy)=6\frac{1}{M}(\frac{\partial{N}}{\partial{x}}-\frac{\partial{M}}{\partial{y}})=6 is a is a constant function , which may be regarded as function of y i,e ϕ(y),\phi(y), then the integration factor of given differential equation is


I.F.=e6.dy=e6yI.F.=e^{\int 6.dy}=e^{6y}


Multiplying by I.F. ,the given differential equation becomes

e6y(xy2x2)dx+e6y(3x2y2+x2y2x3+y2)=0..........(1)e^{6y}(xy^2-x^2)dx+e^{6y}(3x^2y^2+x^2y-2x^3+y^2)=0..........(1)

Which can be shown to be exact.

Therefore the required solution is

y=const.e6y(xy2x2)dx+e6yy2dy=I1+I2(say)\int_{y=const.}e^{6y}(xy^2-x^2)dx+\int e^{6y}y^2 dy=I_1+I_2(say)


Where I1=y=const.e6y(xy2x2)dx,I2=e6yy2dyI_1=\int_{y=const.} e^{6y}(xy^2-x^2)dx ,I_2=\int e^{6y}y^2dy


I1=e6y(y2x22x33).I_1=e^{6y}(\frac{y^2x^2}{2}-\frac{x^3}{3}).

I2=y2e6ydy(dy2dy(e6ydy))dyI_2=y^2\int e^{6y}dy-\int (\frac{dy^2}{dy} (\int e^{6y}dy))dy by using by part.


=y2e6y62y×e6y6dy=\frac{y^2e^{6y}}{6}-\int 2y×\frac{e^{6y}}{6}dy


=y2e6y613ye6ydy+13(dydy(e6ydy))dy=\frac{y^2e^{6y}}{6}-\frac{1}{3}y\int e^{6y}dy+\frac{1}{3}\int( \frac{dy}{dy}(\int e^{6y}dy))dy by using by parts.


=y2e6y6118ye6y+1108e6y=\frac{y^2e^{6y}}{6}-\frac{1}{18}ye^{6y} +\frac{1}{108}e^{6y} .


Therefore the required solution of the given differential equation is

=I1+I2=I_1+I_2

=e6y(y2x22x33)+y2e6y6118ye6y+1108e6y+k=e^{6y}(\frac{y^2x^2}{2}-\frac{x^3}{3})+\frac{y^2e^{6y}}{6}-\frac{1}{18}ye^{6y}+\frac{1}{108}e^{6y}+k

Where k is the integration constant.



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