If the solution y=y(x) of the differential eq | vedclass

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(C) Given the differential equation: $(x^4+2x^3+3x^2+2x+2)dy = (2x^2+2x+3)dx$. Separating the variables,we get: $dy = \frac{2x^2+2x+3}{x^4+2x^3+3x^2+2

Vedclass · Accepted Answer

$\frac{\pi}{4}$

Vedclass · Answer

(C) Given the differential equation: $(x^4+2x^3+3x^2+2x+2)dy = (2x^2+2x+3)dx$. Separating the variables,we get: $dy = \frac{2x^2+2x+3}{x^4+2x^3+3x^2+2x+2}dx$. Factorizing the denominator: $x^4+2x^3+3x^2+2x+2 = (x^2+1)(x^2+2x+2)$. Using partial fractions: $\frac{2x^2+2x+3}{(x^2+1)(x^2+2x+2)} = \frac{1}{x^2+1} + \frac{1}{x^2+2x+2}$. Integrating both sides: $y = \int \frac{1}{x^2+1}dx + \int \frac{1}{(x+1)^2+1}dx$. $y = 	an^{-1}(x) + 	an^{-1}(x+1) + C$. Given $y(-1) = -\frac{\pi}{4}$: $-\frac{\pi}{4} = 	an^{-1}(-1) + 	an^{-1}(0) + C$. $-\frac{\pi}{4} = -\frac{\pi}{4} + 0 + C \Rightarrow C = 0$. Thus,$y(x) = 	an^{-1}(x) + 	an^{-1}(x+1)$. For $y(0)$: $y(0) = 	an^{-1}(0) + 	an^{-1}(1) = 0 + \frac{\pi}{4} = \frac{\pi}{4}$.

If the solution $y=y(x)$ of the differential equation $(x^4+2x^3+3x^2+2x+2)dy-(2x^2+2x+3)dx=0$ satisfies $y(-1)=-\frac{\pi}{4}$,then $y(0)$ is equal to:

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