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Exercise 4 · Q2

Q.Find the general solution of the differential equation: x5dydx=−y5x^5\frac{dy}{dx}=-y^5

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Separable: dyy5=−dxx5\dfrac{dy}{y^5}=-\dfrac{dx}{x^5} integrates to 1x4+1y4=C\dfrac{1}{x^4}+\dfrac{1}{y^4}=C.

Variables-separable; use ∫xn dx=xn+1n+1\displaystyle\int x^{n}\,dx=\frac{x^{n+1}}{n+1} with n=−5n=-5.

Steps

  1. Given:

x5dydx=−y5.x^5\frac{dy}{dx}=-y^5.

  1. Separate variables:

dyy5=−dxx5  ⇒  y−5 dy=−x−5 dx.\frac{dy}{y^5}=-\frac{dx}{x^5}\;\Rightarrow\;y^{-5}\,dy=-x^{-5}\,dx.

  1. Integrate both sides:

∫y−5 dy=−∫x−5 dx.\int y^{-5}\,dy=-\int x^{-5}\,dx.

y−4−4=−x−4−4+C1  ⇒  −14y4=14x4+C1.\frac{y^{-4}}{-4}=-\frac{x^{-4}}{-4}+C_1\;\Rightarrow\;-\frac{1}{4y^4}=\frac{1}{4x^4}+C_1.

  1. Multiply through by −4-4:

1y4=−1x4−4C1.\frac{1}{y^4}=-\frac{1}{x^4}-4C_1.

  1. Move terms together and write C=−4C1C=-4C_1:

1x4+1y4=C.\frac{1}{x^4}+\frac{1}{y^4}=C.

✓Final answer

1x4+1y4=C\dfrac{1}{x^{4}}+\dfrac{1}{y^{4}}=C (equivalently x−4+y−4=Cx^{-4}+y^{-4}=C).

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