prudencemwaura
Answered

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05* Find, for y> 0, the general solution of the differential equation dy/dx=xy.
Inlyl=1/2x^2+c
Inlyl=1/2x^2-c
Inlyl=-1/2x^2-c
Inly|=-1/2x^2+cā€‹

Sagot :

LammettHash

The ODE is separable.

[tex]\dfrac{dy}{dx} = xy \iff \dfrac{dy}y = x\,dx[/tex]

Integrate both sides to get

[tex]\displaystyle \int\frac{dy}y = \int x\,dx[/tex]

[tex]\boxed{\ln|y| = \dfrac12 x^2 + C}[/tex]

But notice that replacing the constant [tex]C[/tex] with [tex]-C[/tex] doesn't affect the solution, since its derivative would recover the same ODE as before.

[tex]\ln|y| = \dfrac12 x^2 - C \implies \dfrac1y \dfrac{dy}{dx} = x \implies \dfrac{dy}{dx} = xy[/tex]

so either of the first two answers are technically correct.

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