Solve the differential equation d y d x +y x=y^2 x.

Question

Solve the differential equation $\frac{d y}{d x}+y \tan x=y^2$ $\sec x$.

✓

Answer

The given differential equation is
$
\begin{array}{r}
\frac{d y}{d x}+y \tan x=y^2 \sec x \\
\Rightarrow \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y} \tan x=\sec x
\end{array}
$
Let $\frac{1}{y}=z \Rightarrow-\frac{1}{y^2} \frac{d y}{d x}=\frac{d z}{d x}$
Substituting
$
\begin{aligned}
-\frac{d z}{d x}+(\tan x) z & =\sec x \\
- & -\frac{d z}{d x}-(\tan x) z=\sec x
\end{aligned}
$
This is a linear differential equation comparing which with $\frac{d z}{d x}+ P z= Q$
$
\begin{aligned}
P & =-\tan x, Q=-\sec x \\
\text { I.F. }=e^{\int P d x} & =e^{\int(-\tan x) d x} \\
& =e^{-\int \tan x d x}=e^{-\log (\sec x)} \\
& =e^{-\log (\sec x)}=\cos x
\end{aligned}
$
Hence the solution of differential equation (3) is
$
\begin{aligned}
z(IF) & =\int(IF) Q d x+c \\
\Rightarrow \quad z(\cos x) & =\int(-\cos x) \cdot(-\sec x) d x+c \\
& =-\int 1 d x+c \\
\Rightarrow \quad z(\cos x) & =-x+c \\
\Rightarrow \quad \frac{1}{y} \cos x & =-x+c \\
\Rightarrow \quad \frac{\cos x}{y}+x & =c
\end{aligned}
$
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