First slide

Methods of solving first order first degree differential equations

Question

If y₁ and y₂ are two solution to the differential equation $\frac{d y}{d x} + p (x) y = Q (x)$ and $y = y_{1} + c k$ then k =

Difficult

A

$y_{1} + 2 y_{2}$
B

$y_{1} - 2 y_{2}$
C

$y_{1} - y_{2}$
D

$2 y_{1} y_{2}$

Solution

$\begin{array}{l} y_{1} y_{2} are two solution of \frac{d y}{d x} + p (x) y = Q (x) \to (1) \\ Then \frac{d y_{1}}{d x} + p (x) y_{1} = Q (x) \to (2) \\ And \frac{d y_{2}}{d x} + p (x) y_{2} = Q (x) \to (3) \\ E q (1) - E q (2) \Rightarrow \frac{d (y - y_{1})}{d x} + p (x) (y - y_{1}) = 0 \to (4) \\ E q (2) - E q (3) ⇒ \frac{d (y_{1} - y_{2})}{d x} + p (x) (y_{1} - y_{2}) = 0 \to (5) \\ \frac{E q (4)}{E q (5)} \Rightarrow \frac{\frac{d}{d x} (y_{1} - y_{2})}{\frac{d}{d x} (y_{1} - y_{2})} = \frac{y - y_{1}}{y_{1} - y_{2}} \end{array}$
$\begin{array}{l} \int \frac{\frac{d}{d x} (y - y_{1})}{y - y_{1}} = \int \frac{\frac{d}{d x} (y_{1} - y_{2})}{y_{1} - y_{2}} \\ \Rightarrow \ln |y - y_{1}| = \ln |y_{1} - y_{2}| + \ln c \\ \Rightarrow y - y_{1} = c (y_{1} - y_{2}) \\ \Rightarrow y = y_{1} + c (y_{1} - y_{2}) \end{array}$

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