Solve dy/dx+y/x=x^2y^6 [Bernoulli Differential Equation]

In this page, we solve dy/dx+y/x=x^2y^6. This is a Bernoulli’s differential equation. The general solution of dy/dx+y/x=x^2y^6 is 5x³y⁵ + Cx⁵y⁵ = 2 where C denotes an arbitrary constant of integrals.

Let us now solve the Bernoulli’s differential equation dy/dx+y/x=x²y⁶.

Solution of dy/dx+y/x=x^2y^6

Question: Find the solution of $\dfrac{dy}{dx}+\dfrac{y}{x}=x^2y^6$.

Answer:

$\dfrac{dy}{dx}+\dfrac{y}{x}=x^2y^6$ …(I)

The given differential equation is in Bernoulli’s form: $\dfrac{dy}{dx}+P(x)=Q(x)y^n$ where P(x), Q(x) are functions of x, and n is an integer.

Here P(x) = 1/x, Q(x) = x², n=6.

Step 1:

Divide both sides of (I) by y⁶. So we have that

$y^{-6}\dfrac{dy}{dx}+\dfrac{1}{x} y^{-5}=x^2$ …(II)

Step 2:

Put y^-5 = z.

Differentiating both sides, we have

-5y^-6 $\dfrac{dy}{dx}$ = $\dfrac{dz}{dx}$

⇒ y^-6 $\dfrac{dy}{dx}$ = $-\dfrac{1}{5}\dfrac{dz}{dx}$

Step 3:

Substituting the above values from step 3 in Equation (II), we obtain that

$-\dfrac{1}{5}\dfrac{dz}{dx}+\dfrac{1}{x} z=x^2$

⇒ $\dfrac{dz}{dx}-\dfrac{5}{x} z= -5x^2$ …(IV)

It is of the form $\dfrac{dz}{dx}+P(x) z= Q(x)$ with P(x) = -5/x and Q(x) = -5x².

This is a linear first order differential equation in z.

Its integrating factor (IF) = $e^{\int P(x)~dx}$ $= e^{-5\ln x} = e^{\ln x^{-5}} = x^{-5}$.

Step 4:

∴ The solution of (IV) is given by

z × IF = $\displaystyle \int$ (IF × Q(x)) dx + K

⇒ $y^{-5} x^{-5} = \displaystyle \int x^{-5} \times (-5x^2)dx$ + K

⇒ $\dfrac{1}{x^5y^5} = \displaystyle \int x^{-5} \times (-5x^2)dx$ + K

⇒ $\dfrac{1}{x^5y^5} = -5\displaystyle \int x^{-3} dx$ + K

⇒ $\dfrac{1}{x^5y^5} = -5 \dfrac{x^{-3+1}}{-3+1} dx$ + K

⇒ $\dfrac{1}{x^5y^5} = \dfrac{5}{2x^2} dx$ + K

⇒ $2 = 5x^3y^5 dx + Cx^5y^5$ where C=2K.

Therefore, 5x³y⁵ + Cx⁵y⁵ = 2 is the solution of dy/dx+y/x=x²y⁶, where C is an arbitrary constant of integrations.

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