The solution of the differential equation d y-d x-xx-y-x3x-y3-1 is x-y-k-c ex2-x2-1 then k-23-k4-

Put x + y = z 1+dy/dx=dz/dx Equation becomes, dz/dx+xz=x^3z^3 z^ 3dz/dx+xz^ 2=x^3 Put z^ 2=t.Then the equation becomes dt/dx 2xt= 2x^3 Solution is 1/(x+y)^2=1+x ...

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Standard XII

Mathematics

Algebra of Derivatives

The solution ...

Question

The solution of the differential equation $\frac{d y}{d x} + x (x + y) = x^{3} (x + y)^{3} - 1$ is $(x + y)^{- k} = c e^{x^{2}} + x^{2} + 1$ then $\frac{(k + 2)^{3}}{k^{4}} =$ ___

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Solution

Put x + y = z
$1 + \frac{d y}{d x} = \frac{d z}{d x}$
Equation becomes, $\frac{d z}{d x} + x z = x^{3} z^{3}$
$z^{- 3} \frac{d z}{d x} + x z^{- 2} = x^{3}$
Put $z^{- 2} = t$ .Then the equation becomes $\frac{d t}{d x} - 2 x t = - 2 x^{3}$
Solution is $\frac{1}{(x + y)^{2}} = 1 + x^{2} + c e^{x^{2}}$ . Hence $K = 2$

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The solution of the differential equation dydx+x(x+y)=x3(x+y)3−1 is (x+y)−k=cex2+x2+1 then (k+2)3k4= ___

Put x + y = z 1+dydx=dzdx Equation becomes, dzdx+xz=x3z3 z−3dzdx+xz−2=x3 Put z−2=t.Then the equation becomes dtdx−2xt=−2x3 Solution is 1(x+y)2=1+x2+cex2. Hence K=2

The solution of the differential equation $\frac{d y}{d x} + x (x + y) = x^{3} (x + y)^{3} - 1$ is $(x + y)^{- k} = c e^{x^{2}} + x^{2} + 1$ then $\frac{(k + 2)^{3}}{k^{4}} =$ ___