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# Solve differential equation 6(y-x^2)dx+xdy=0 # Solve differential equation 6(y-x^2)dx+xdy=0

Question
First order differential equations asked 2020-11-23
Solve differential equation $$6(y-x^2)dx+xdy=0$$

## Answers (1) 2020-11-24
$$6(y-x^2)dx+xdy=0$$
$$(6(y-x^2)dx)/(xdx)+(xdy)/(xdx)=0$$
$$(6(y-x^2)dx)/(xdx)+(xdy)/(xdx)=0$$
$$6((y-x^2)/x)+dy/dx=0$$ $$dy/dx+(6/x)y=6x$$ (1)
Now, equation (1) is first order linear differential equation
We know that solution of first order linear differential equation is given by
$$y.(I.F.)= int (I.F.)6xdx$$ (2)x
where, $$I.F.= e^(int 6/x dx)$$
$$I.F. = e^(6 int dx/x)$$ $$( :' int dx/x= ln x+c )$$
$$= e^(6(ln x))$$
$$= e^(ln x^6)$$
$$= x^6$$
Now, solving equation(2) by using I.F.
$$y(x^6)= int x^6(6x)dx$$
$$= 6 int x^7dx$$ $$( :' int x^n dx= (x^(n+1))/(n+1)+c$$
$$= 6(x^8/8)+c$$
$$= (3x^8)/4+c$$ (3)
Now, using y(1)= 1 in equation (3)
$$1(1)^6= (3(1)^8)/4+c$$
$$1= 3/4+c$$
$$c= 1-3/4$$
$$c= (4-3)/4$$
$$c= 1/4$$
Now, putting value of c in equation (3) $$yx^6= (3x^8)/4+1/4$$
Hence, solution of given differential equation is $$yx^6=(3x^8)/4+1/4$$

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