Differential Equations

First Order Equations

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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

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

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

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