Solve the differential equation. xy' − 2y = x², x > 0

Solution:

The standard form of differential equations is more easy and quick to solve.

Go through the step-by-step solution to get the final answer.

Given equation, xy' - 2y = x²

Step 1) Divide both the sides by x

⇒ y' - 2y / x = x

Step 2) Compare it with the standard form, y' + P(x)y = Q(x)

We have P(x) = -2 / x , Q(x) = x

Thus the integrating factor, IF = e ^{∫ P(x) dx}

IF = e ^{∫ P(x) dx} = e ^{-2∫1/x dx} = x ^-2

Step 3) Multiplying both sides by x^-2, we get

⇒  y × IF = ∫  (Q × IF) dx + C

⇒ y x^-2 =   ∫  (x ×x^-2 ) dx+ C

^⇒ yx^-2 =   ∫  (x^-1 ) dx+ C

Step 4) Integrating both sides.

^⇒ yx^-2 =  ln|x| + C

⇒ y = x² ln|x| + C x² 

So, the solution of xy' - 2y = x² is y = x² ln|x| + C x².

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Solve the differential equation. xy' − 2y = x², x > 0

Summary:

The integral of  xy' − 2y = x² gives the solution as y = x² ln|x| + cx²