Help with simple differential equations y dy= x dx.?
I am told to find a solution of the differential equations y dy= x dx. What I did is integrate left and right hand side respectively end up with y^2=x^2+2c when its all cleaned up. The solution book says I should get is x^2-y^2=C and I can see how I can get y^2-x^2=C but not x^2-y^2=C like book has.
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answers:
Jeffrey K say: Your answer is correct and so is the book. Multiply your equation by -1. Since c is an arbitrary constant, you can let C = -c and you match the book.
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alex say: integrate both side
y² = x² + C
or
y² + C= x²
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Φ² = Φ+1 say: y² - x² = C₀
-(y² - x²) = -C₀
-y² + x² = C₁
x² - y² = C₁
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Mike G say: It just means that the Cs will have opposite signs.
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Como say: :-
Integrate
y²/2 = x²/2 + C
y² = x² + 2C
y² = x² + K
y = [ x² + K ] ^(1/2)
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Captain Matticus, LandPiratesInc say: They're essentially the same, because you don't have a specific value for C

1) y^2 = x^2 + C
2) y^2 - x^2 = C
3) x^2 - y^2 = C

The C in 2 is not the same as the C in 3. Your math is good, if that's any comfort.
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King Leo say: .
it’s just algebra somehow

y dy = x dx
y dy - x dx = 0
∫ y dy - ∫ x dx = ∫ 0
½y² - ½x² = c
y² - x² = 2c
y² - x² = C
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y dy = x dx
0 = x dx - y dy
∫ 0 = ∫ x dx - ∫ y dy
c = ½x² - ½y²
2c = x² - y²
C = x² - y²
x² - y² = C
—————
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