- #1
Pengwuino
Gold Member
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- 20
Given a differential equation with the form:
[tex]\frac{{dy}}{{dx}} + P(x)y = Q(x)y^n [/tex]
and using the substitution [tex]v = y^{1 - n}[/tex]
I attempted to prove that it transforms into
[tex]\frac{{dv}}{{dx}} + (1 - n)P(x)v = (1 - n)Q(x)[/tex]
Here’s the proof, did I do it correctly? I got the write answer so I assume I did :D
[tex]\begin{array}{l}
y = v^{ - 1 + n} \\
\frac{{dv}}{{dx}} = \frac{{dv}}{{dy}}\frac{{dy}}{{dx}} \\
\frac{{dv}}{{dy}} = (1 - n)y^{ - n} \frac{{dy}}{{dx}} \\
\frac{{y^n }}{{(1 - n)}}\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}} \\
\frac{{y^n }}{{(1 - n)}}\frac{{dv}}{{dx}} + P(x)y = Q(x)y^n \\
\frac{{dv}}{{dx}} + (1 - n)P(x)\frac{y}{{y^n }} = (1 - n)Q(x) \\
\frac{{dv}}{{dx}} + (1 - n)P(x)y^{1 - n} = (1 - n)Q(x) \\
\frac{{dv}}{{dx}} + (1 - n)P(x)v = (1 - n)Q(x) \\
\end{array}
[/tex]
[tex]\frac{{dy}}{{dx}} + P(x)y = Q(x)y^n [/tex]
and using the substitution [tex]v = y^{1 - n}[/tex]
I attempted to prove that it transforms into
[tex]\frac{{dv}}{{dx}} + (1 - n)P(x)v = (1 - n)Q(x)[/tex]
Here’s the proof, did I do it correctly? I got the write answer so I assume I did :D
[tex]\begin{array}{l}
y = v^{ - 1 + n} \\
\frac{{dv}}{{dx}} = \frac{{dv}}{{dy}}\frac{{dy}}{{dx}} \\
\frac{{dv}}{{dy}} = (1 - n)y^{ - n} \frac{{dy}}{{dx}} \\
\frac{{y^n }}{{(1 - n)}}\frac{{dv}}{{dx}} = \frac{{dy}}{{dx}} \\
\frac{{y^n }}{{(1 - n)}}\frac{{dv}}{{dx}} + P(x)y = Q(x)y^n \\
\frac{{dv}}{{dx}} + (1 - n)P(x)\frac{y}{{y^n }} = (1 - n)Q(x) \\
\frac{{dv}}{{dx}} + (1 - n)P(x)y^{1 - n} = (1 - n)Q(x) \\
\frac{{dv}}{{dx}} + (1 - n)P(x)v = (1 - n)Q(x) \\
\end{array}
[/tex]
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