Answer to Question #109878 in Differential Equations for Yiit

Question

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

Answer 1

¡) Ans:

The given differential equation is

$y sin2x dx=(1+y^2+cos^2x)dy$

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

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

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

$$\frac{\partial{M}}{\partial{y}}=\frac{\partial{ysin2x}}{\partial{y}}=sin2x$$ $$\frac{\partial{N}}{\partial{x}}=\frac{\partial{-(1+y^2+cos^2x)}}{\partial{x}}=-(2cosx×-sinx)$$

$=2sinxcos =sin2x$

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

$\therefore \$ The solution of the given differential equation is

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

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

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

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

Where "c" is the integration constant.

¡¡)ans :

The given differential equation is

$(xy^2-x^2)dx+(3x^2y^2+x^2y-2x^3+y^2)dy=0$

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

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

$\therefore \frac{\partial{M}}{\partial{y}}=2xy \ \text{and} \ \frac{\partial{N}}{\partial{x}}=6xy^2+2xy-6x^2$

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

Therefore ,the given differential equation is not exact.

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

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

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

$e^{6y}(xy^2-x^2)dx+e^{6y}(3x^2y^2+x^2y-2x^3+y^2)=0..........(1)$

Which can be shown to be exact.

Therefore the required solution is

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

Where $I_1=\int_{y=const.} e^{6y}(xy^2-x^2)dx ,I_2=\int e^{6y}y^2dy$

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

$I_2=y^2\int e^{6y}dy-\int (\frac{dy^2}{dy} (\int e^{6y}dy))dy$ by using by part.

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

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

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

Therefore the required solution of the given differential equation is

$=I_1+I_2$

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

Where k is the integration constant.