Question
Solve the following differential equation
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
$\sqrt{1-\text{x}^4}\text{dy}=\text{x dx}$
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$\vec{\text{a}}=4\hat{\text{i}}-3\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=2\hat{\text{i}}-4\hat{\text{j}}+5\hat{\text{k}},\vec{\text{c}}=\hat{\text{i}}-\hat{\text{j}}$ from a right triangle.