# Bernoulli Equation

## Solved Problems

Click or tap a problem to see the solution.

### 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

$z' = \left( {{y^{ - 3}}} \right)^\prime = - 3{y^{ - 4}}y' = - \frac{{3y'}}{{{y^4}}}.$

Multiply both sides of the original equation by $$\left( { - 3} \right)$$ and divide by $${{y^4}}:$$

$y' + y\cot x = {y^4}\sin x,\;\; \Rightarrow - \frac{{3y'}}{{{y^4}}} - \frac{{3\cot x}}{{{y^3}}} = - 3\sin x.$

Notice that in dividing by $${{y^4}},$$ we have lost the solution $$y = 0.$$ Rewriting the last equation in terms of $$z,$$ we get

$z' - 3\cot x \cdot z = - 3\sin x.$

This differential equation is linear, so we can solve it using the integrating factor:

$u\left( x \right) = {e^{\int {\left( { - 3} \right)\cot xdx} }} = {e^{ - 3\int {\cot xdx} }} = {e^{ - 3\int {\frac{{\cos xdx}}{{\sin x}}} }} = {e^{ - 3\int {\frac{{d\left( {\sin x} \right)}}{{\sin x}}} }} = {e^{ - 3\ln \left| {\sin x} \right|}} = {e^{\ln \frac{1}{{{{\left| {\sin x} \right|}^3}}}}} = \frac{1}{{{{\left| {\sin x} \right|}^3}}}.$

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 $$u\left( x \right):$$

$z' \cdot \frac{1}{{{{\sin }^3}x}} - 3\cot x \cdot z \cdot \frac{1}{{{{\sin }^3}z}} = z'\frac{1}{{{{\sin }^3}z}} - \frac{{3z\cos x}}{{{{\sin }^4}x}} = \left( {z\frac{1}{{{{\sin }^3}x}}} \right)^\prime.$

Hence, the general solution of the linear differential equation for $$z\left( x \right)$$ can be presented in the form:

$z = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {\frac{1}{{{{\sin }^3}x}}\left( { - 3\sin x} \right)dx} + C}}{{\frac{1}{{{{\sin }^3}x}}}} = \frac{{ - 3\int {\frac{{dx}}{{{{\sin }^2}x}}} + C}}{{\frac{1}{{{{\sin }^3}x}}}} = \left( {3\cot x + C} \right){\sin ^3}x.$

Since $$z = {y^{ - 3}},$$ we obtain the following solutions of the given Bernoulli equation:

$\frac{1}{{{y^3}}} = \left( {3\cot x + C} \right){\sin ^3}x,\;\; y = 0.$

### 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

$z' = {\left( {\sqrt y } \right)^\prime } = \frac{{y'}}{{2\sqrt y }}.$

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

$y' + \frac{{2y}}{x} = 2x\sqrt y ,\;\; \Rightarrow \frac{{y'}}{{2\sqrt y }} + \frac{{\cancel{2}y}}{{\cancel{2}x\sqrt y }} = \frac{{\cancel{2}x\sqrt y }}{{\cancel{2}\sqrt y }},\;\; \Rightarrow \frac{{y'}}{{2\sqrt y }} + \frac{{\sqrt y }}{x} = x.$

Replacing $$y$$ with $$z,$$ we get

$z' + \frac{z}{x} = x.$

We obtain a simple linear equation for the function $$z\left( x \right).$$ The integrating factor here is

$u\left( x \right) = {e^{\int {\frac{1}{x}dx} }} = {e^{\ln \left| x \right|}} = \left| x \right|.$

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 $$u\left( x \right):$$

$z' \cdot x + \frac{z}{x} \cdot x = z'x + z = {\left( {zx} \right)^\prime }.$

Then the general solution of the linear differential equation is given by

$z = \frac{{\int {u\left( x \right)f\left( x \right)dx} + C}}{{u\left( x \right)}} = \frac{{\int {x \cdot xdx} + C}}{x} = \frac{{\int {{x^2}dx} + C}}{x} = \frac{{\frac{{{x^3}}}{3} + C}}{x}.$

Returning to the original function $$y\left( x \right),$$ we get the solution in the implicit form:

$\sqrt y = \frac{{\frac{{{x^3}}}{3} + C}}{x}\;\;\text{or}\;\;x\sqrt y = \frac{{{x^3}}}{3} + C.$

Thus, the full answer is written as follows:

$x\sqrt y = \frac{{{x^3}}}{3} + C,\;\; y = 0.$

### 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:

$4xyy' = {y^2} + {x^2},\;\; \Rightarrow \frac{{4xyy'}}{{4xy}} - \frac{{{y^2}}}{{4xy}} = \frac{{{x^2}}}{{4xy}},\;\; \Rightarrow y' - \frac{y}{{4x}} = \frac{x}{{4y}}.$

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 $$2y:$$

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

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

$z' - \frac{z}{{2x}} = \frac{x}{2}.$

Calculate the integrating factor:

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

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 $$u\left( x \right):$$

$\left( {z^\prime - \frac{z}{{2x}}} \right)u\left( x \right) = z^\prime \cdot \frac{1}{{\sqrt x }} - \frac{z}{{2x}} \cdot \frac{1}{{\sqrt x }} = z^\prime \cdot \frac{1}{{\sqrt x }} - z \cdot \frac{1}{{2{x^{\frac{3}{2}}}}} = z^\prime \cdot \frac{1}{{\sqrt x }} - z \cdot \frac{{{x^{ - \frac{3}{2}}}}}{2} = z^\prime \cdot \frac{1}{{\sqrt x }} + z \cdot \left( {{x^{ - \frac{1}{2}}}} \right)^\prime = z^\prime \cdot \frac{1}{{\sqrt x }} + z \cdot \left( {\frac{1}{{\sqrt x }}} \right)^\prime = \left( {z \cdot \frac{1}{{\sqrt x }}} \right)^\prime$

Find the general solution of the linear equation:

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

Taking into account that $$z = {y^2},$$ we obtain the following solution:

$y = \pm \sqrt {\frac{{{x^2}}}{3} + C\sqrt x } .$

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,

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

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

So the solution of the IVP is given by the function

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