Practicing Success
Practicing Success
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The differential equation $\frac{d y}{d x}+P y=Q y^n$, $n>2$ can be reduced to linear form by substituting |
$z=y^{n-1}$ $z=y^n$ $z=y^{n+1}$ $z=y^{1-n}$ |
$z=y^{1-n}$ |
We have, $\frac{1}{y^n} \frac{d y}{d x}+P \frac{1}{y^{n-1}}=Q$ It reduces to linear form by substituting $\frac{1}{y^{n-1}}=z$ |
