Question:
If
\[
2x-y+c\log|x-2y-4|=k
\]
is the general solution of
\[
\frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2},
\]
then \(c=\)
If
\[
2x-y+c\log|x-2y-4|=k
\]
is the general solution of
\[
\frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2},
\]
then \(c=\)
Show Hint
When a differential equation contains expressions like \(x-2y\), use substitution \(u=x-2y\) to reduce it into a separable form.
Updated On: Jun 22, 2026
- \(4\)
- \(2\)
- \(3\)
- \(-4\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use substitution.
Given differential equation is \[ \frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2} \] Let \[ u=x-2y \] Then, \[ \frac{du}{dx}=1-2\frac{dy}{dx} \] Now, \[ \frac{dy}{dx}=\frac{2u-5}{u+2} \] Therefore, \[ \frac{du}{dx} = 1-2\left(\frac{2u-5}{u+2}\right) \] \[ = \frac{u+2-4u+10}{u+2} \] \[ = \frac{-3u+12}{u+2} \] \[ = \frac{-3(u-4)}{u+2} \]
Step 2: Separate the variables.
\[ \frac{dx}{du} = -\frac{u+2}{3(u-4)} \] Now, \[ \frac{u+2}{u-4} = \frac{u-4+6}{u-4} = 1+\frac{6}{u-4} \] So, \[ dx = -\frac{1}{3}\left(1+\frac{6}{u-4}\right)du \]
Step 3: Integrate both sides.
\[ x = -\frac{1}{3}\int \left(1+\frac{6}{u-4}\right)du \] \[ x = -\frac{u}{3}-2\log|u-4|+C \] Multiplying by \(3\), \[ 3x=-u-6\log|u-4|+C \] \[ 3x+u+6\log|u-4|=C \]
Step 4: Substitute back \(u=x-2y\).
Since \[ u=x-2y, \] we get \[ 3x+x-2y+6\log|x-2y-4|=C \] \[ 4x-2y+6\log|x-2y-4|=C \] Dividing by \(2\), \[ 2x-y+3\log|x-2y-4|=k \]
Step 5: Compare with the given solution.
Given solution is \[ 2x-y+c\log|x-2y-4|=k \] Comparing both equations, \[ c=3 \]
Step 6: Final conclusion.
Hence, \[ \boxed{3} \]
Given differential equation is \[ \frac{dy}{dx}=\frac{2x-4y-5}{x-2y+2} \] Let \[ u=x-2y \] Then, \[ \frac{du}{dx}=1-2\frac{dy}{dx} \] Now, \[ \frac{dy}{dx}=\frac{2u-5}{u+2} \] Therefore, \[ \frac{du}{dx} = 1-2\left(\frac{2u-5}{u+2}\right) \] \[ = \frac{u+2-4u+10}{u+2} \] \[ = \frac{-3u+12}{u+2} \] \[ = \frac{-3(u-4)}{u+2} \]
Step 2: Separate the variables.
\[ \frac{dx}{du} = -\frac{u+2}{3(u-4)} \] Now, \[ \frac{u+2}{u-4} = \frac{u-4+6}{u-4} = 1+\frac{6}{u-4} \] So, \[ dx = -\frac{1}{3}\left(1+\frac{6}{u-4}\right)du \]
Step 3: Integrate both sides.
\[ x = -\frac{1}{3}\int \left(1+\frac{6}{u-4}\right)du \] \[ x = -\frac{u}{3}-2\log|u-4|+C \] Multiplying by \(3\), \[ 3x=-u-6\log|u-4|+C \] \[ 3x+u+6\log|u-4|=C \]
Step 4: Substitute back \(u=x-2y\).
Since \[ u=x-2y, \] we get \[ 3x+x-2y+6\log|x-2y-4|=C \] \[ 4x-2y+6\log|x-2y-4|=C \] Dividing by \(2\), \[ 2x-y+3\log|x-2y-4|=k \]
Step 5: Compare with the given solution.
Given solution is \[ 2x-y+c\log|x-2y-4|=k \] Comparing both equations, \[ c=3 \]
Step 6: Final conclusion.
Hence, \[ \boxed{3} \]
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