y+x2=dydx has the solution

y+x2=dydx⇒ dydx−y=x2I.F=e∫−1 dx=e−xSolution is y.(I.F)=∫x2.(I.F) dx⇒ y.e−x=∫x2e−x dx=x2∫e−xdx−∫(2x)∫e−x dx dx=−x2e−x+2∫x e−xdx=−x2e−x+2(x(−e−x)−∫−e−xdx)=−x2e−x−2xe−x−2e−x+c⇒ y+x2+2x+2=cex

y+x2=dydx has the solution

y+x2=dydx⇒ dydx−y=x2I.F=e∫−1 dx=e−xSolution is y.(I.F)=∫x2.(I.F) dx⇒ y.e−x=∫x2e−x dx=x2∫e−xdx−∫(2x)∫e−x dx dx=−x2e−x+2∫x e−xdx=−x2e−x+2(x(−e−x)−∫−e−xdx)=−x2e−x−2xe−x−2e−x+c⇒ y+x2+2x+2=cex

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$y + x^{2} = \frac{dy}{dx}$ has the solution

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$y + x^{2} + 2 x + 2 = {ce}^{x}$

$y + x + 2 x^{2} + 2 = {ce}^{x}$

$y + x + x^{2} + 2 = {ce}^{2 x}$

$y^{2} + x + x^{2} + 2 = {ce}^{2 x}$

answer is A.

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Detailed Solution

$\begin{array}{l} y + x^{2} = \frac{d y}{d x} \\ \Rightarrow \frac{d y}{d x} - y = x^{2} \\ I . F = e^{\int - 1 d x} = e^{- x} \\ S o l u t i o n i s y . (I . F) = \int x^{2} . (I . F) d x \\ \Rightarrow y . e^{- x} = \int x^{2} e^{- x} d x \\ = x^{2} \int e^{- x} d x - \int (2 x) \int e^{- x} d x d x \end{array}$

$\begin{array}{l} = - x^{2} e^{- x} + 2 \int x e^{- x} d x \\ = - x^{2} e^{- x} + 2 (x (- e^{- x}) - \int - e^{- x} d x) \\ = - x^{2} e^{- x} - 2 x e^{- x} - 2 e^{- x} + c \\ \Rightarrow y + x^{2} + 2 x + 2 = c e^{x} \end{array}$