Question: "Solvingy+xy′=a(1+xy)y(1a)=−a"

esbagoar7kh

Answered question

2022-04-22

Solving

y + x y^{'} = a (1 + x y)

y (\frac{1}{a}) = - a

Answer & Explanation

Felicity Carter

Beginner2022-04-23Added 16 answers

Step 1
Well, it is not hard to see that we rewrite your DE in the following form:
1) $y^{'} (x) + \frac{(1 - a x) y (x)}{x} = \frac{a}{x}$
Now, let:
2) $μ (x) := \exp {\int \frac{1 - a x}{x} d x} = x \exp (- a x)$
When we multiply both sides by $μ (x)$ , use the fact that
$\exp (- a x) (1 - a x) = \frac{d}{d x} (x \exp (- a x))$
and apply the reverse product rule we end up with:
3) $\frac{d}{d x} (x \exp (- a x) y (x)) = a \exp (- a x)$
Integrate both sides with respect to x:
4) $x \exp (- a x) y (x) = C - \exp (- a x)$
Dividing both sides by $μ (x) :$
5) $y (x) = \frac{C \exp (a x) - 1}{x}$
So, we can solve for C:
6) $y (\frac{1}{a}) = \frac{C \exp (a \cdot \frac{1}{a}) - 1}{\frac{1}{a}} = a (C e - 1) = - a \leftrightarrow C = \frac{2}{e}$
So, we end up with:
7) $y (x) = \frac{\frac{2}{e} \cdot \exp (a x) - 1}{x} = \frac{2 \exp (a x) - e}{e x} = \frac{2 \exp (a x)}{e x} - \frac{1}{x}$

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