Bernoulli's equation is one of the most famous nonlinear differential equations of the first order. It is written in the form

Where a(x) And b(x) are continuous functions. If m= 0, then Bernoulli's equation becomes a linear differential equation. In the case when m= 1, the equation becomes a separable equation. In general, when m≠ 0.1, Bernoulli's equation is reduced to a linear differential equation using the substitution

New differential equation for the function z(x) has the form

and can be solved using the methods described on the page First-order linear differential equations.

BERNOULI METHOD.

The equation under consideration can be solved by Bernoulli's method. To do this, we look for a solution to the original equation in the form of a product of two functions: where u, v- functions from x. Differentiate: Substitute into the original equation (1): (2) As v Let’s take any non-zero solution to the equation: (3) Equation (3) is an equation with separable variables. After we found its particular solution v = v(x), substitute it into (2). Since it satisfies equation (3), the expression in parentheses becomes zero. We get: This is also a separable equation. We find its general solution, and with it the solution to the original equation y = uv.

64. Equation in total differentials. Integrating factor. Solution methods

First order differential equation of the form

called equation in total differentials, if its left side represents the total differential of some function, i.e.

Theorem. In order for equation (1) to be an equation in total differentials, it is necessary and sufficient that in some simply connected domain of change of variables the condition is satisfied

The general integral of equation (1) has the form or

Example 1. Solve differential equation.

Solution. Let's check that this equation is a total differential equation:

so that is condition (2) is satisfied. Thus, this equation is an equation in total differentials and

therefore, where is still an undefined function.

Integrating, we get . The partial derivative of the found function must be equal to, which gives from where so that Thus,.

General integral of the original differential equation.

When integrating some differential equations, the terms can be grouped in such a way that easily integrable combinations are obtained.

65. Ordinary differential linear equations of higher orders: homogeneous and inhomogeneous. Linear differential operator, its properties (with proof).

Linear differential operator and its properties. The set of functions having on the interval ( a , b ) no less n derivatives, forms a linear space. Consider the operator L n (y ), which displays the function y (x ), having derivatives, into a function having k - n derivatives.

Linear differential equations of 1st order
and Bernoulli's equation

A first-order linear differential equation is an equation that is linear with respect to an unknown function and its derivative. It looks like

\frac(dy)(dx)+p(x)y=q(x),

where p(x) and q(x) are given functions of x, continuous in the region in which equation (1) needs to be integrated.

If q(x)\equiv0 , then equation (1) is called linear homogeneous. It is a separable equation and has a general solution

y=C\exp\!\left(-\int(p(x))\,dx\right)\!,

The general solution to the inhomogeneous equation can be found method of variation of an arbitrary constant, which consists in the fact that the solution to equation (1) is sought in the form

y=C(x)\exp\!\left(-\int(p(x))\,dx\right), where C(x) is a new unknown function of x.

Example 1. Solve the equation y"+2xy=2xe^(-x^2).

Solution. Let's use the constant variation method. Consider the homogeneous equation y"+2xy=0, corresponding to this inhomogeneous equation. This is an equation with separable variables. Its general solution has the form y=Ce^(-x^2) .

We look for a general solution to the inhomogeneous equation in the form y=C(x)e^(-x^2), where C(x) is an unknown function of x. Substituting, we get C"(x)=2x, whence C(x)=x^2+C. So, the general solution of the inhomogeneous equation will be y=(x^2+C)e^(-x^2), where C is the constant of integration.

Comment. It may turn out that the differential equation is linear in x as a function of y. The normal form of such an equation is

\frac(dx)(dy)+r(y)x=\varphi(y).

Example 2. Solve the equation \frac(dy)(dx)=\frac(1)(x\cos(y)+\sin2y).

Solution. This equation is linear if we consider x as a function of y:

\frac(dx)(dy)-x\cos(y)=\sin(2y).

We use the method of variation of an arbitrary constant. First we solve the corresponding homogeneous equation

\frac(dx)(dy)-x\cos(y)=0,

which is an equation with separable variables. Its general solution has the form x=Ce^(\sin(y)),~C=\text(const).

We look for a general solution to the equation in the form , where C(y) is an unknown function of y. Substituting, we get

C"(y)e^(\sin(y))=\sin2y or C"(y)=e^(-\sin(y))\sin2y.

From here, integrating by parts, we have

\begin(aligned)C(y)&=\int(e^(-\sin(y))\sin2y)\,dy=2\int(e^(-\sin(y))\cos(y) \sin(y))\,dy=2\int\sin(y)\,d(-e^(-\sin(y)))=\\ &=-2\sin(y)\,e^ (-\sin(y))+2\int(e^(-\sin(y))\cos(y))\,dy=C-2(\sin(y)+1)e^(-\ sin(y)),\end(aligned)

C(y)=-2e^(-\sin(y))(1+\sin(y))+C.

Substituting this equation into x=C(y)e^(\sin(y)), we obtain a general solution to the original equation, and therefore to this equation:

x=Ce^(\sin(y))-2(1+\sin(y))

The original equation can also be integrated as follows. We believe

y=u(x)v(x),

where u(x) and v(x) are unknown functions of x, one of which, for example v(x), can be chosen arbitrarily.

Substituting y=u(x)v(x) into , after transformation we get

vu"+(pv+v")u=q(x).

Determining v(x) from the condition v"+pv=0, we then find from vu"+(pv+v")u=q(x) function u(x) and, consequently, the solution y=uv to the equation \frac(dy)(dx)+p(x)y=q(x). As v(x) we can take any frequent solution of the equation v"+pv=0,~v\not\equiv0.

Example 3. Solve the Cauchy problem: x(x-1)y"+y=x^2(2x-1),~y|_(x=2)=4.

Solution. We are looking for a general solution to the equation in the form y=u(x)v(x) ; we have y"=u"v+uv". Substituting the expression for y and y" into the original equation, we will have

x(x-1)(u"v+uv")+uv=x^2(2x-1) or x(x-1)vu"+u=x^2(2x-1)

We find the function v=v(x) from the condition x(x-1)v"+v=0. Taking any particular solution of the last equation, for example v=\frac(x)(x-1) and substituting it, we get the equation u"=2x-1, from which we find the function u(x)=x^2-x+C. Therefore, the general solution to the equation x(x-1)y"+y=x^2(2x-1) will

y=uv=(x^2-x+C)\frac(x)(x-1), or y=\frac(Cx)(x-1)+x^2.

Using the initial condition y|_(x=2)=4, we obtain the equation for finding C 4=\frac(2C)(2-1)+2^2, from where C=0 ; so the solution to the stated Cauchy problem will be the function y=x^2.

Example 4. It is known that there is a relationship between current i and electromotive force E in a circuit having resistance R and self-inductance L E=Ri+L\frac(di)(dt), where R and L are constants. If we consider E a function of time t, we obtain a linear inhomogeneous equation for the current strength i:

\frac(di)(dt)+\frac(R)(L)i(t)=\frac(E(t))(L).

Find the current strength i(t) for the case when E=E_0=\text(const) and i(0)=I_0 .

Solution. We have \frac(di)(dt)+\frac(R)(L)i(t)=\frac(E_0)(L),~i(0)=I_0. The general solution of this equation has the form i(t)=\frac(E_0)(R)+Ce^(-(R/L)t). Using the initial condition (13), we obtain from C=I_0-\frac(E_0)(R), so the desired solution will be

i(t)=\frac(E_0)(R)+\left(I_0-\frac(E_0)(R)\right)\!e^(-(R/L)t).

This shows that at t\to+\infty the current strength i(t) tends to a constant value \frac(E_0)(R) .

Example 5. A family C_\alpha of integral curves of the linear inhomogeneous equation y"+p(x)y=q(x) is given.

Show that the tangents at the corresponding points to the curves C_\alpha defined by the linear equation intersect at one point (Fig. 13).

Solution. Consider the tangent to any curve C_\alpha at the point M(x,y). The equation of the tangent at the point M(x,y) has the form

\eta-q(x)(\xi-x)=y, where \xi,\eta are the current coordinates of the tangent point.

By definition, at the corresponding points x is constant and y is variable. Taking any two tangents to the lines C_\alpha at the corresponding points, for the coordinates of the point S of their intersection, we obtain

\xi=x+\frac(1)(p(x)), \quad \eta=x+\frac(q(x))(p(x)).

This shows that all tangents to the curves C_\alpha at the corresponding points (x is fixed) intersect at the same point

S\!\left(x+\frac(1)(p(x));\,x+\frac(q(x))(p(x))\right).

Eliminating the argument x in the system, we obtain the equation of the locus of points S\colon f(\xi,\eta)=0.

Example 6. Find the solution to the equation y"-y=\cos(x)-\sin(x), satisfying the condition: y is limited at y\to+\infty .

Solution. The general solution to this equation is y=Ce^x+\sin(x) . Any solution to the equation obtained from the general solution for C\ne0 will be unbounded, since for x\to+\infty the function \sin(x) is bounded and e^x\to+\infty . It follows that this equation has a unique solution y=\sin(x) , bounded at x\to+\infty , which is obtained from the general solution at C=0 .

Bernoulli's equation

Bernoulli's differential equation looks like

\frac(dy)(dx)+p(x)y=q(x)y^n, where n\ne0;1 (for n=0 and n=1 this equation is linear).

Using variable replacement z=\frac(1)(y^(n-1)) Bernoulli's equation is reduced to a linear equation and integrated as a linear one.

Example 7. Solve Bernoulli's equation y"-xy=-xy^3.

Solution. Divide both sides of the equation by y^3:

\frac(y")(y^3)-\frac(x)(y^2)=-x

Making a variable change \frac(1)(y^2)=z\Rightarrow-\frac(2y")(y^3)=z", where \frac(y")(y^3)=-\frac(z")(2). After substitution, the last equation turns into a linear equation

-\frac(z")(2)-xz=-x or z"+2xz=2x, the general solution of which is z=1+Ce^(-x^2).

From here we obtain the general integral of this equation

\frac(1)(y^2)=1+Ce^(-x^2) or y^2(1+Ce^(-x^2))=1.

Comment. Bernoulli's equation can also be integrated by the method of variation of a constant, like a linear equation, and using the substitution y(x)=u(x)v(x) .

Example 8. Solve Bernoulli's equation xy"+y=y^2\ln(x). .

Solution. Let us apply the method of variation of an arbitrary constant. The general solution of the corresponding homogeneous equation xy"+y=0 has the form y=\frac(C)(x). We look for the general solution of the equation in the form y=\frac(C(x))(x) , where C(x) - new unknown function. Substituting into the original equation, we will have

C"(x)=C^2(x)\frac(\ln(x))(x^2).

To find the function C(x), we obtain an equation with separable variables, from which, by separating the variables and integrating, we find

\frac(1)(C(x))=\frac(\ln(x))(x)+\frac(1)(x)+C~\Rightarrow~C(x)=\frac(x)( 1+Cx+\ln(x)).

So, the general solution to the original equation y=\frac(1)(1+Cx+\ln(x)).

Some nonlinear first-order equations can be reduced to linear equations or Bernoulli equations using a successfully found change of variables.

Example 9. Solve the equation y"+\sin(y)+x\cos(y)+x=0.

Solution. Let us write this equation in the form y"+2\sin\frac(y)(2)\cos\frac(y)(2)+2x\cos^2\frac(y)(2)=0..

Dividing both sides of the equation by 2\cos^2\frac(y)(2), we get \frac(y")(2\cos^2\dfrac(y)(2))+\operatorname(tg)\frac(y)(2)+x=0.

Replacement \operatorname(tg)\frac(y)(2)=z\Rightarrow\frac(dz)(dx)=\frac(y")(\cos^2\dfrac(y)(2)) reduces this equation to linear \frac(dz)(dx)+z=-x, the general solution of which is z=1-x+Ce^(-x) .

Replacing z by its expression in terms of y, we obtain the general integral of this equation \operatorname(tg)\frac(y)(2)=1-x+Ce^(-x).

In some equations, the desired function y(x) may be under the integral sign. In these cases, it is sometimes possible to reduce this equation to a differential equation by differentiation.

Example 10. Solve the equation x\int\limits_(x)^(0)y(t)\,dt=(x+1)\int\limits_(0)^(x)ty(t)\,dt,~x>0.

Solution. Differentiating both sides of this equation with respect to x, we get

\int\limits_(0)^(x)y(t)\,dt+xy(x)=\int\limits_(0)^(x)ty(t)\,dt+x(x+1)y (x) or \int\limits_(0)^(x)y(t)\,dx=\int\limits_(0)^(x)ty(t)\,dt+x^2y(x).

Differentiating again with respect to x, we will have a linear homogeneous equation with respect to y(x)\colon

y(x)=xy(x)+x^2y"(x)+2xy(x) or x^2y"(x)+(3x-1)y(x)=0.

Separating the variables and integrating, we find y=\frac(C)(x^3)e^(-1/x). This solution, as can be easily verified, satisfies the original equation.

A first-order linear differential equation is an equation that is linear with respect to an unknown function and its derivative. It looks like

\frac(dy)(dx)+p(x)y=q(x),

where p(x) and q(x) are given functions of x, continuous in the region in which equation (1) needs to be integrated.

If q(x)\equiv0 , then equation (1) is called linear homogeneous. It is a separable equation and has a general solution

Y=C\exp\!\left(-\int(p(x))\,dx\right)\!,

The general solution to the inhomogeneous equation can be found method of variation of an arbitrary constant, which consists in the fact that the solution to equation (1) is sought in the form

Y=C(x)\exp\!\left(-\int(p(x))\,dx\right), where C(x) is a new unknown function of x.

Example 1. Solve the equation y"+2xy=2xe^(-x^2) .

Solution. Let's use the constant variation method. Consider the homogeneous equation y"+2xy=0, corresponding to this inhomogeneous equation. This is an equation with separable variables. Its general solution has the form y=Ce^(-x^2) .

We look for a general solution to the inhomogeneous equation in the form y=C(x)e^(-x^2), where C(x) is an unknown function of x. Substituting, we get C"(x)=2x, whence C(x)=x^2+C. So, the general solution of the inhomogeneous equation will be y=(x^2+C)e^(-x^2) , where C - constant of integration.

Comment. It may turn out that the differential equation is linear in x as a function of y. The normal form of such an equation is

\frac(dx)(dy)+r(y)x=\varphi(y).

Example 2. Solve the equation \frac(dy)(dx)=\frac(1)(x\cos(y)+\sin2y).

Solution. This equation is linear if we consider x as a function of y:

\frac(dx)(dy)-x\cos(y)=\sin(2y).

We use the method of variation of an arbitrary constant. First we solve the corresponding homogeneous equation

\frac(dx)(dy)-x\cos(y)=0,

which is an equation with separable variables. Its general solution has the form x=Ce^(\sin(y)),~C=\text(const).

We look for a general solution to the equation in the form x=C(y)e^(\sin(y)), where C(y) is an unknown function of y. Substituting, we get

C"(y)e^(\sin(y))=\sin2y or C"(y)=e^(-\sin(y))\sin2y.

From here, integrating by parts, we have

\begin(aligned)C(y)&=\int(e^(-\sin(y))\sin2y)\,dy=2\int(e^(-\sin(y))\cos(y) \sin(y))\,dy=2\int\sin(y)\,d(-e^(-\sin(y)))=\\ &=-2\sin(y)\,e^ (-\sin(y))+2\int(e^(-\sin(y))\cos(y))\,dy=C-2(\sin(y)+1)e^(-\ sin(y)),\end(aligned)

So,

C(y)=-2e^(-\sin(y))(1+\sin(y))+C.

Substituting this equation into x=C(y)e^(\sin(y)) , we obtain a general solution to the original equation, and therefore to this equation:

X=Ce^(\sin(y))-2(1+\sin(y))

The original equation can also be integrated as follows. We believe

Y=u(x)v(x),

where u(x) and v(x) are unknown functions of x, one of which, for example v(x), can be chosen arbitrarily.

Substituting y=u(x)v(x) into , after transformation we get

Vu"+(pv+v")u=q(x).

Determining v(x) from the condition v"+pv=0, we then find from vu"+(pv+v")u=q(x) the function u(x) and, consequently, the solution y=uv of the equation \frac(dy)(dx)+p(x)y=q(x). As v(x) we can take any frequent solution of the equation v"+pv=0,~v\not\equiv0.

Example 3. Solve the Cauchy problem: x(x-1)y"+y=x^2(2x-1),~y|_(x=2)=4.

Solution. We are looking for a general solution to the equation in the form y=u(x)v(x) ; we have y"=u"v+uv". Substituting the expression for y and y" into the original equation, we will have

X(x-1)(u"v+uv")+uv=x^2(2x-1) or x(x-1)vu"+u=x^2(2x-1)

We find the function v=v(x) from the condition x(x-1)v"+v=0. Taking any particular solution of the last equation, for example v=\frac(x)(x-1) and substituting it, we get the equation u"=2x-1, from which we find the function u(x)=x^2-x+C. Therefore, the general solution to the equation x(x-1)y"+y=x^2(2x-1) will

Y=uv=(x^2-x+C)\frac(x)(x-1), or y=\frac(Cx)(x-1)+x^2.

Using the initial condition y|_(x=2)=4, we obtain the equation for finding C 4=\frac(2C)(2-1)+2^2, from where C=0 ; so the solution to the stated Cauchy problem will be the function y=x^2.

Example 4. It is known that there is a relationship between current i and electromotive force E in a circuit having resistance R and self-inductance L E=Ri+L\frac(di)(dt), where R and L are constants. If we consider E a function of time t, we obtain a linear inhomogeneous equation for the current strength i:

\frac(di)(dt)+\frac(R)(L)i(t)=\frac(E(t))(L).

Find the current strength i(t) for the case when E=E_0=\text(const) and i(0)=I_0 .

Solution. We have \frac(di)(dt)+\frac(R)(L)i(t)=\frac(E_0)(L),~i(0)=I_0. The general solution of this equation has the form i(t)=\frac(E_0)(R)+Ce^(-(R/L)t). Using the initial condition (13), we obtain from C=I_0-\frac(E_0)(R), so the desired solution will be

I(t)=\frac(E_0)(R)+\left(I_0-\frac(E_0)(R)\right)\!e^(-(R/L)t).

This shows that at t\to+\infty the current strength i(t) tends to a constant value \frac(E_0)(R) .

Example 5. A family C_\alpha of integral curves of the linear inhomogeneous equation y"+p(x)y=q(x) is given.

Show that the tangents at the corresponding points to the curves C_\alpha defined by the linear equation intersect at one point (Fig. 13).

Solution. Consider the tangent to any curve C_\alpha at the point M(x,y). The equation of the tangent at the point M(x,y) has the form

\eta-q(x)(\xi-x)=y, where \xi,\eta are the current coordinates of the tangent point.

By definition, at the corresponding points x is constant and y is variable. Taking any two tangents to the lines C_\alpha at the corresponding points, for the coordinates of the point S of their intersection, we obtain

\xi=x+\frac(1)(p(x)), \quad \eta=x+\frac(q(x))(p(x)).

This shows that all tangents to the curves C_\alpha at the corresponding points ( x is fixed) intersect at the same point

S\!\left(x+\frac(1)(p(x));\,x+\frac(q(x))(p(x))\right).

Eliminating the argument x in the system, we obtain the equation of the locus of points S\colon f(\xi,\eta)=0.

Example 6. Find the solution to the equation y"-y=\cos(x)-\sin(x), satisfying the condition: y is limited at y\to+\infty .

Solution. The general solution to this equation is y=Ce^x+\sin(x) . Any solution to the equation obtained from the general solution for C\ne0 will be unbounded, since for x\to+\infty the function \sin(x) is bounded and e^x\to+\infty . It follows that this equation has a unique solution y=\sin(x) , bounded at x\to+\infty , which is obtained from the general solution at C=0 .

Bernoulli's equation

Bernoulli's differential equation looks like

\frac(dy)(dx)+p(x)y=q(x)y^n, where n\ne0;1 (for n=0 and n=1 this equation is linear).

Using variable replacement z=\frac(1)(y^(n-1)) Bernoulli's equation is reduced to a linear equation and integrated as a linear one.

Example 7. Solve Bernoulli's equation y"-xy=-xy^3.

Solution. Divide both sides of the equation by y^3:

\frac(y")(y^3)-\frac(x)(y^2)=-x

Making a variable change \frac(1)(y^2)=z\Rightarrow-\frac(2y")(y^3)=z", where \frac(y")(y^3)=-\frac(z")(2). After substitution, the last equation turns into a linear equation

-\frac(z")(2)-xz=-x or z"+2xz=2x, the general solution of which is z=1+Ce^(-x^2).

From here we obtain the general integral of this equation

\frac(1)(y^2)=1+Ce^(-x^2) or y^2(1+Ce^(-x^2))=1.

Comment. Bernoulli's equation can also be integrated by the method of variation of a constant, like a linear equation, and using the substitution y(x)=u(x)v(x) .

Example 8. Solve Bernoulli's equation xy"+y=y^2\ln(x). .

Solution. Let us apply the method of variation of an arbitrary constant. The general solution of the corresponding homogeneous equation xy"+y=0 has the form y=\frac(C)(x). We look for the general solution of the equation in the form y=\frac(C(x))(x) , where C(x) - new unknown function. Substituting into the original equation, we will have

C"(x)=C^2(x)\frac(\ln(x))(x^2).

To find the function C(x), we obtain an equation with separable variables, from which, by separating the variables and integrating, we find

\frac(1)(C(x))=\frac(\ln(x))(x)+\frac(1)(x)+C~\Rightarrow~C(x)=\frac(x)( 1+Cx+\ln(x)).

So, the general solution to the original equation y=\frac(1)(1+Cx+\ln(x)).

Some nonlinear first-order equations can be reduced to linear equations or Bernoulli equations using a successfully found change of variables.

Example 9. Solve the equation y"+\sin(y)+x\cos(y)+x=0.

Solution. Let us write this equation in the form y"+2\sin\frac(y)(2)\cos\frac(y)(2)+2x\cos^2\frac(y)(2)=0..

Dividing both sides of the equation by 2\cos^2\frac(y)(2), we get \frac(y")(2\cos^2\dfrac(y)(2))+\operatorname(tg)\frac(y)(2)+x=0.

Replacement \operatorname(tg)\frac(y)(2)=z\Rightarrow\frac(dz)(dx)=\frac(y")(\cos^2\dfrac(y)(2)) reduces this equation to linear \frac(dz)(dx)+z=-x, the general solution of which is z=1-x+Ce^(-x) .

Replacing z by its expression in terms of y, we obtain the general integral of this equation \operatorname(tg)\frac(y)(2)=1-x+Ce^(-x).

In some equations, the desired function y(x) may be under the integral sign. In these cases, it is sometimes possible to reduce this equation to a differential equation by differentiation.

Example 10. Solve the equation x\int\limits_(x)^(0)y(t)\,dt=(x+1)\int\limits_(0)^(x)ty(t)\,dt,~x>0.

Solution. Differentiating both sides of this equation with respect to x, we get

\int\limits_(0)^(x)y(t)\,dt+xy(x)=\int\limits_(0)^(x)ty(t)\,dt+x(x+1)y (x) or Source of information

Characteristics of Bernoulli's equation

Definition 1

First order differential equation having the standard form $y"+P\left(x\right)\cdot y=Q\left(x\right)\cdot y^(n)$, where $P\left(x\right )$ and $Q\left(x\right)$ are continuous functions, and $n$ is a certain number, called the Jacob Bernoulli differential equation.

In this case, restrictions are imposed on the number $n$:

• $n\ne 0$, since at $n = 0$ the differential equation is linear inhomogeneous, and some other special solution method is not needed in this case;
• $n\ne 1$, since if we have one as $n$, the differential equation is a linear homogeneous one, the solution method of which is also known.

In addition, the trivial solution of the Bernoulli differential equation $y=0$ is not specifically considered.

The differential equation of mathematician Jacob Bernoulli should not be confused with Bernoulli's law, named after his nephew's uncle, known as Daniel Bernoulli.

Note 1

Daniel Bernoulli is a physicist, the most famous pattern he found is to describe the relationship between the speed of fluid flow and pressure. Bernoulli's law is also applicable for laminar gas flows. It is generally used in hydraulics and fluid dynamics.

Solution of the Bernoulli equation by reduction to a linear inhomogeneous

The main method for solving the Bernoulli differential equation is that through transformations it is reduced to a linear inhomogeneous one. These transformations are as follows:

1. We multiply the equation by the number $y^(-n) $ and get $y^(-n) \cdot y"+P\left(x\right)\cdot y^(1-n) =Q\left(x\ right)$.
2. We apply the replacement $z=y^(1-n) $ and differentiate this equality as a complex power function; we get $z"=\left(1-n\right)\cdot y^(-n) \cdot y"$, whence $\frac(z")(1-n) =y^(-n) \cdot y"$.
3. We substitute the values ​​$y^(1-n) $ and $y^(-n) \cdot y"$ into this differential equation and get $\frac(z")(1-n) +P\left(x\right )\cdot z=Q\left(x\right)$ or $z"+\left(1-n\right)\cdot P\left(x\right)\cdot z=\left(1-n\right )\cdot Q\left(x\right)$.

The resulting differential equation is linear inhomogeneous with respect to the function $z$, which we solve as follows:

1. We calculate the integral $I_(1) =\int \left(1-n\right)\cdot P\left(x\right)\cdot dx $, write a particular solution in the form $v\left(x\right)=e ^(-I_(1) ) $, we perform simplifying transformations and choose the simplest non-zero option for $v\left(x\right)$.
2. We calculate the integral $I_(2) =\int \frac(\left(1-n\right)\cdot Q\left(x\right))(v\left(x\right)) \cdot dx $, after which we write the expression in the form $u\left(x,C\right)=I_(2) +C$.
3. We write the general solution of a linear inhomogeneous differential equation in the form $z=u\left(x,C\right)\cdot v\left(x\right)$.
4. We return to the function $y$, replacing $z$ with $y^(1-n)$, and, if necessary, perform simplifying transformations.

Example:

Find the general solution to the differential equation $\frac(dy)(dx) +\frac(y)(x) =y^(2) \cdot \left(4-x^(2) \right)$. Write down a particular solution that satisfies the initial condition $y=1$ for $x=1$.

In this case, we have the Bernoulli differential equation presented in standard form.

In this case, $n=2$, $P\left(x\right)=\frac(1)(x) $, $Q\left(x\right)=4-x^(2) $.

We present it in the form regarding the replacement $z$:

$z"+\left(1-2\right)\cdot \frac(1)(x) \cdot z=\left(1-2\right)\cdot \left(4-x^(2) \right )$ or $z"-\frac(1)(x) \cdot z=-\left(4-x^(2) \right)$.

The resulting differential equation is linear inhomogeneous with respect to the function $z$, which we solve using the method described above.

We calculate the integral $I_(1) =\int \left(1-n\right)\cdot P\left(x\right)\cdot dx $.

We have $I_(1) =\int \left(1-2\right)\cdot \frac(1)(x) \cdot dx =-\ln \left|x\right|$.

We write a particular solution in the form $v\left(x\right)=e^(-I_(1) ) $ and perform simplifying transformations: $v\left(x\right)=e^(\ln \left|x\ right|)$; $\ln v\left(x\right)=\ln \left|x\right|$; $v\left(x\right)=\left|x\right|$.

For $v\left(x\right)$ we choose the simplest non-zero option: $v\left(x\right)=x$.

We calculate the integral $I_(2) =\int \frac(\left(1-n\right)\cdot Q\left(x\right))(v\left(x\right)) \cdot dx $.

We write the expression in the form $u\left(x,C\right)=I_(2) +C$, that is, $u\left(x,C\right)=\frac(x^(2) )(2) -4\cdot \ln \left|x\right|+C$.

We finally write down the general solution of the linear inhomogeneous differential equation for the function $z$ in the form $z=u\left(x,C\right)\cdot v\left(x\right)$, that is, $z=\frac(x^ (3) )(2) -4\cdot x\cdot \ln \left|x\right|+C\cdot x$.

Now we return to the function $y$, replacing $z$ with $y^(1-n)$:

$y^(1-2) =\frac(x^(3) )(2) -4\cdot x\cdot \ln \left|x\right|+C\cdot x$ or $\frac(1) (y) =\frac(x^(3) )(2) -4\cdot x\cdot \ln \left|x\right|+C\cdot x$.

This is the general solution of this Bernoulli differential equation, written in implicit form.

To find a particular solution, we use this initial condition $y=1$ for $x=1$:

Therefore, the partial solution has the form: $\frac(1)(y) =\frac(x^(3) )(2) -4\cdot x\cdot \ln \left|x\right|+\frac(x )(2) $.

Solving Bernoulli's differential equation by substitution method

The second possible solution to Bernoulli's equation is the substitution method.

Example:

Find the general solution to the differential equation $y"+\frac(y)(x) =y^(2) \cdot \left(4-x^(2) \right)$ by substitution method.

We apply the substitution $y=u\cdot v$.

After differentiation we get:

We find the function $v\left(x\right)$ from the equation $v"+\frac(v)(x) =0$; to do this, we move the second term to the right side.

We get:

$\frac(dv)(dx) =-\frac(v)(x) $;

separate the variables $\frac(dv)(v) =-\frac(dx)(x) $;

integrate $\ln \left|v\right|=-\ln \left|x\right|$, whence $v=\frac(1)(x) $.

The function $u\left(x\right)$ is found from the equation $u"\cdot \frac(1)(x) =u^(2) \cdot \frac(1)(x^(2) ) \cdot \ left(4-x^(2) \right)$, which takes into account $v=\frac(1)(x) $ and $v"+\frac(v)(x) =0$.

After simple transformations we get: $u"=u^(2) \cdot \frac(1)(x) \cdot \left(4-x^(2) \right)$.

We separate the variables: $\frac(du)(u^(2) ) =\frac(1)(x) \cdot \left(4-x^(2) \right)\cdot dx$.

Let's integrate: $-\frac(1)(u) =4\cdot \ln \left|x\right|-\frac(x^(2) )(2) +C$ or $\frac(1)(u ) =\frac(x^(2) )(2) -4\cdot \ln \left|x\right|+C$.

Let's return to the old variable. We take into account that $y=u\cdot v$ or $y=u\cdot \frac(1)(x) $, whence $u=x\cdot y$.

We obtain the general solution to this differential equation: $\frac(1)(y) =\frac(x^(3) )(2) -4\cdot x\cdot \ln \left|x\right|+C\cdot x $.