Bernoulli Equation
Solved Problems
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Example 3
Solve the equation \[y' + y\cot x = {y^4}\sin x.\]
Example 4
Find all solutions of the differential equation \[y' + \frac{{2y}}{x} = 2x\sqrt y .\]
Example 5
Find the solution of the differential equation \[4xyy' = {y^2} + {x^2},\] satisfying the initial condition \(y\left( 1 \right) = 1.\)
Example 3.
Solve the equation \[y' + y\cot x = {y^4}\sin x.\]
Solution.
This is a Bernoulli equation with the parameter \(m = 4.\) Therefore, we make the substitution \(z = {y^{1 - m}} = {y^{ - 3}}.\) The derivative is given by
Multiply both sides of the original equation by \(\left( { - 3} \right)\) and divide by \({{y^4}}:\)
Notice that in dividing by \({{y^4}},\) we have lost the solution \(y = 0.\) Rewriting the last equation in terms of \(z,\) we get
This differential equation is linear, so we can solve it using the integrating factor:
We can take the function \(u\left( x \right) = \frac{1}{{{{\sin }^3}x}}\) as the integrating factor. In fact, the left side of the equation becomes the derivative of the product \(z\left( x \right)u\left( x \right)\) after multiplying by \(u\left( x \right):\)
Hence, the general solution of the linear differential equation for \(z\left( x \right)\) can be presented in the form:
Since \(z = {y^{ - 3}},\) we obtain the following solutions of the given Bernoulli equation:
Example 4.
Find all solutions of the differential equation \[y' + \frac{{2y}}{x} = 2x\sqrt y .\]
Solution.
This equation is also a Bernoulli equation with the fractional parameter \(m = {\frac{1}{2}}.\) It can be reduced to the linear equation by making the replacement \(z = {y^{1 - m}} \) \(= \sqrt y .\) The derivative of the new function \(z\left( x \right)\) is given by
Divide the original Bernoulli equation by \({2\sqrt y }.\) Like in other examples on this page, the root \(y = 0\) is also the trivial solution of the differential equation. So we have
Replacing \(y\) with \(z,\) we get
We obtain a simple linear equation for the function \(z\left( x \right).\) The integrating factor here is
We choose the function \(u\left( x \right) = x.\) One can check that the left side of the equation becomes the derivative of the product \(z\left( x \right)u\left( x \right)\) after multiplying by \(u\left( x \right):\)
Then the general solution of the linear differential equation is given by
Returning to the original function \(y\left( x \right),\) we get the solution in the implicit form:
Thus, the full answer is written as follows:
Example 5.
Find the solution of the differential equation \[4xyy' = {y^2} + {x^2},\] satisfying the initial condition \(y\left( 1 \right) = 1.\)
Solution.
First we should check whether this differential equation is a Bernoulli equation:
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 \(2y:\)
By replacing \(y\) with \(z,\) we can convert the Bernoulli equation into the linear differential equation:
Calculate the integrating factor:
Let's choose the function \(u\left( x \right) = \frac{1}{{\sqrt x }}\) and make sure that the left side of the equation becomes the derivative of the product \(z\left( x \right)u\left( x \right)\) after multiplying by \(u\left( x \right):\)
Find the general solution of the linear equation:
Taking into account that \(z = {y^2},\) we obtain the following solution:
Now we determine the value of the constant \(C\) that matches the initial condition \(y\left( 1 \right) = 1.\) We see that only solution with the positive sign satisfies this condition. Hence,
This gives: \(C = \frac{2}{3} .\)
So the solution of the IVP is given by the function