Question

Solve and find particular solution of the first order non-linear differential equation of Bernoulli’s type with given condition.

 

Answer 1

Solution:

the first-order non-linear differential equation of Bernoulli’s type is

$$\dfrac{dy}{dx}+P(x)y=Q(x)y^n.$$

where p(x) and q(x) are continuous functions on the interval we’re working on and n is a real number. Differential equations in this form are called Bernoulli Equations.

Example:

Find the solution of the differential equation $4 x y y ′ = y ^2 + x ^2$, satisfying the initial condition y(1)=1.

Solution:

First, we should check whether this differential equation is a Bernoulli equation:

$$4xyy'=y^2+x^2,\;\rightarrow\;\dfrac{4xyy'}{4xy}-\dfrac{y^2}{4xy}=\dfrac{x^2}{4xy},\;\rightarrow\;y'-\dfrac{y}{4x}=\dfrac{x}{4y}$$

As it can be seen, we have a Bernoulli equation with the parameter m=−1. Hence, we can make the substitution $z=y^{1-m}=y^2.$ The derivative of the function is $z'=2yy'.$ Next, we multiply both sides of the differential equation by 2y:

$$2yy'-\dfrac{2y^2}{4x}=\dfrac{2xy}{4y}, \;\rightarrow\;2yy'-\dfrac{y^2}{2x}=\dfrac{x}{2}.$$

By replacing y with z, we can convert the Bernoulli equation into the linear differential equation:

$$z'=\dfrac{z}{2x}=\dfrac{x}{2}.$$

Calculate the integrating factor:

$$u(x)=e^{\int(-\frac{1}{2x})dx}=e^{{-\frac{1}{2}\int-\frac{dx}x}}=e^{-\frac{1}{2}ln|x|}=\dfrac{1}{\sqrt{|x|}}$$

Let`s choose the function $u(x)=\dfrac{1}{\sqrt{x}}$ and make sure that the left side of the equation becomes the derivative of the product z(x) u(x) after multiplying by u(x):

$(z'-\dfrac{z}{2x})u(x)=z'\;\cdot\dfrac{1}{\sqrt{x}}-\dfrac{z}{2x}\cdot \dfrac{1}{\sqrt{x}}=z'\cdot\dfrac{1}{\sqrt{x}}-z\cdot\dfrac{1}{2x^{\frac{3}{2}}}=z'\cdot\dfrac{1}{\sqrt{x}}-z\cdot\dfrac{x^{-\frac{3}{2}}}{2}=z'\cdot\dfrac{1}{\sqrt{x}}+z\cdot(x^{\frac{1}{2}})'=z'\cdot\dfrac{1}{\sqrt{x}}+z\cdot(\dfrac{1}{\sqrt{x}})'=(z\cdot\dfrac{1}{\sqrt{x}})'$

Find the general solution of the linear equation:

$z=\dfrac{\int u(x)f(x)dx+C}{u(x)}=\dfrac{\int \dfrac{1}{\sqrt{x}}\cdot\frac{x}{2}dx+C}{\dfrac{1}{\sqrt{x}}}=\dfrac{\frac{1}{2}\int \sqrt{x}dx+C}{\dfrac{1}{\sqrt{x}}}=\sqrt{x}[\dfrac{1}{2}\cdot \dfrac{2x^{\frac{3}{2}}}{3}+C]=\dfrac{x^2}{3}+C\sqrt{x}.$

Taking into account that z=y², we obtain the following solution:

$$y=\plusmn\sqrt{\dfrac{x^2}{3}+C\sqrt{x}}.$$

Now we determine the value of the constant C

 that matches the initial condition y(1)=1. We see that only a solution with a positive sign satisfies this condition. Hence,

$$y=\sqrt{\dfrac{1^2}{3}+C\sqrt{1}}=\sqrt{\dfrac{1}{3}+C}=1.$$

This gives: $C=\dfrac{2}{3}.$

$$y=\sqrt{\dfrac{x^2}{3}+\dfrac{2\sqrt{x}}{3}}.$$