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Question Number 128179 by bounhome last updated on 05/Jan/21

$$showmethaty=x^2+4x-6isthesolutionof$$$$y^{″}-y^{\prime }+2(x+1)=0$$

Answered by bemath last updated on 05/Jan/21

$$\mathrm{y}={\mathrm{x}}^2+4\mathrm{x}-6\to \{\begin{array}{l}{\mathrm{y}}^{\prime }=2\mathrm{x}+4\\ {\mathrm{y}}^{″}=2\end{array}$$$$(\ast )2-(2\mathrm{x}+4)+2(\mathrm{x}+1)=$$$$2-2\mathrm{x}-4+2\mathrm{x}+2=0$$

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