Linear Equations. Miscellaneous Problems

Solve the equation $xy''+2y'+xy=0$, given the particular solution $xy_{1}=\sin x$.

Solve the equation $(2x-x^{2})y''+(x^{2}-2)y'+2(1-x)y=0$, given the particular solution $y_{1}=x^{2}$.

Solve the equation $x^{2}(\ln x-1)y''-xy'+y=0$, given the particular solution $y_{1}=x$.

Solve the equation $xy''-(1+x)y'+y=0$, given the particular solution $y_{1}=1+x$.

Solve the equation $y''+(\tg x-2\ctg x)y'+2\ctg^{2}x\cdot y=0$, given the particular solution $y_{1}=\sin x$.

Solve the equation $y''\sin^{3}x=4y\sin 3x$, given the particular solution $y_{1}=\sin^{4}x$.

Solve the equation $x(1-x)^{2}y''=2y$, given the particular integral $(1-x)y_{1}=x$.

Solve the equation $(1+x^{2})y''+xy'-n^{2}y=0$, given the particular solution $y_{1}=\left(x+\sqrt{x^{2}+1}\right)^{n}$.

Solve the equation $(2x+1)y''+(4x-2)y'-8y=0$, given the particular solution $y_{1}=e^{mx}$ for a suitable choice of $m$.

Solve the equation $x(x-1)y''-(2x-1)y'+2y=2x^{3}-3x^{2}$, given the particular integral of the corresponding homogeneous equation $z_{1}=x^{2}$.

If a linear equation with rational coefficients has a polynomial as a particular solution, it can easily be found. Thus, if the equation $(1-x^{2})y''-xy'+9y=0$ has a solution in the form of a polynomial $y=Ax^{n}+Bx^{n-1}+\ldots +M$, then after substituting the polynomial into the left-hand side of this equation, the coefficients of the powers of $x$, after collecting like terms, must equal zero. For the coefficient of the highest power of $x$, we get the equality: $-An(n-1)-An+9A=0$, or $A(n^{2}-9)=0$. Since $A\neq 0$ and $n\neq -3$, we have $n=3$. Therefore, if $y$ in the given equation can be a polynomial, it can only be a polynomial of the third degree. Setting $y=Ax^{3}+Bx^{2}+Cx+D$ and substituting into the given equation, we equate to zero the coefficients of the powers of $x$. This gives the equations: $B=0$, $4C=-3A$, $D=0$. Hence $y=A_{1}(4x^{3}-3x)$ is a solution of the given equation.

Integrate the equation $xy''-(x+5)y'+3y=0$, which has a particular solution equal to a polynomial.

Integrate the equation $(x^{2}-1)y''=6y$, which has a particular solution equal to a polynomial.

Integrate the equation $y''-2xy'+4y=0$, which has a particular solution equal to a polynomial.

Integrate the equation $xy''-(x+2)y'+y=0$, which has a particular solution equal to a polynomial.

Show that Legendre's equation $\left((x^{2}-1)y'\right)'-n(n+1)y=0$, where $n$ is a positive integer, has particular solutions of the form $y_{1}=P(x)$ and $y_{2}=P(x)\ln \dfrac {x+1}{x-1}+Q(x)$, where $P(x)$ and $Q(x)$ are polynomials.

For what value of $\mu$ does the equation

have a particular solution in the form of a third-degree polynomial? Find it.

Show that the equation $xy''-(x+p+q)y'+py=0$, where $p$ and $q$ are positive integers, has one particular solution in the form of a polynomial, and another in the form $e^{x}P(x)$, where $P(x)$ is also a polynomial.

Integrate the equation $(x^{2}-x)y''+(2x-3)y'-2y=0$, knowing that its general integral is a rational function (for an equation of the form $a(x)y''+b(x)y'+c(x)y=0$ with polynomial coefficients, the denominator of such an integral can consist only of linear factors that are divisors of the polynomial $a(x)$).

Integrate the equation $(x^{2}-x)y''+(x+1)y'-y=0$, knowing that its general integral is a rational function (see the note to problem 4396).

Integrate the equation $(x^{2}-x)y''+(4x-2)y'+2y=0$, knowing that its general integral is a rational function (see the note to problem 4396).

Integrate the equation $x^{2}(2x+1)(x+1)y''-2x(x^{2}+3x+1)y'+(6x+2)y=0$, knowing that its general integral is a rational function (see the note to problem 4396).

Integrate the equation $(1+x^{2})y''+6xy'+6y=0$, knowing that its general integral is a rational function (see the note to problem 4396).

Integrate the equation $2xy''+y'-2y=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $x^{4}y''+2x^{3}y'+n^{2}y=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $(1+x^{2})^{2}y''+2x(1+x^{2})y'+y=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $(1-x^{2})y''-xy'+n^{2}y=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $y''\ch^{2}2x+y'\sh 4x+\dfrac {m^{2}}{4}y=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $y''\sin x\cos x-y'+m^{2}y\tg x\sin^{2}x=0$, introducing a new independent variable $x=\varphi (t)$, chosen so that the coefficient of $\dfrac {dy}{dt}$ vanishes in the transformed equation.

Integrate the equation $x^{4}y''+2x^{3}y'-4y=\dfrac {1}{x}$, setting $x=\dfrac {1}{t}$.

Integrate the equation $xy''+2y'-xy=e^{x}$, setting $y=\dfrac {u}{x}$.

Integrate the equation $y''+\dfrac {2}{x}y'-a^{2}y=2$, setting $y=\dfrac {u}{x}$.

Integrate the equation $x^{4}y''+c^{2}y=0$, setting $y=xu$, $x=\dfrac {1}{t}$.

Integrate the equation $x^{4}y''-c^{2}y=0$, setting $y=\dfrac {u}{t}$, $x=\dfrac {1}{t}$.

Integrate the equation $\sin^{2}xy''+\sin x\cos xy'-y=0$, by finding a suitable integrating factor.

Solve the equation

by the substitution $x^{2}y''-2xy'+2y=u$. Here $P$ and $Q$ are given functions of $x$.

Integrate the equation $xy''-4y'-xy=0$, applying differentiation to it four times and using the resulting equalities.

Find the general integral of Stokes' equation

where $\beta$ is a constant, by finding a particular solution of the form $y=x^{m}(1-x)^{n}$.

Find the solution of Stokes' equation

such that $y=y'=0$ at $x=0$.