Question

У"+12у'+36у=2cosx-sinx​

Answer 1

Ответ:

y+12y+36y=2cos(x)-sin(x)

49y=2cos(x)-sin(x)

y=$\frac{2 cos(x)}{49}$-$\frac{sin(x)}{49}$

y=$\frac{2 cos(x)-sin(x)}{49}$

y=$\frac{2 cos(x)-sin(x)}{49}$   x∈R

Пошаговое объяснение:x∈R т.е. любое число

Answer 2

To solve the given second-order linear homogeneous differential equation:

$u'' + 12u' + 36u = 2\cos(x) - \sin(x)$

We'll first find the complementary solution (also known as the homogeneous solution) and then look for a particular solution. The general solution will be the sum of the complementary and particular solutions.

Step 1: Find the complementary solution The complementary solution is the solution to the homogeneous equation (right-hand side is zero):

$u'' + 12u' + 36u = 0$

The characteristic equation for this differential equation is:

$r^2 + 12r + 36 = 0$

Solving this quadratic equation for $r$:

$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = 12$, and $c = 36$.

$r = \frac{-12 \pm \sqrt{12^2 - 4 \cdot 1 \cdot 36}}{2 \cdot 1}$

$r = \frac{-12 \pm \sqrt{144 - 144}}{2}$

Since the discriminant is zero, the roots are repeated, and the complementary solution will have the form:

$u_c(x) = (c_1 + c_2x)e^{-6x}$

where $c_1$ and $c_2$ are arbitrary constants.

Step 2: Find the particular solution We are looking for a particular solution to the non-homogeneous part of the equation:

$2\cos(x) - \sin(x)$

We can use the method of undetermined coefficients to find a particular solution. Since the right-hand side contains both cosine and sine terms, we assume a particular solution of the form:

$u_p(x) = A\cos(x) + B\sin(x)$

where $A$ and $B$ are constants to be determined.

Taking the first and second derivatives of $u_p(x)$:

$u'_p(x) = -A\sin(x) + B\cos(x)$

$u''_p(x) = -A\cos(x) - B\sin(x)$

Now, substitute these derivatives and $u_p(x)$ into the original differential equation:

$-A\cos(x) - B\sin(x) + 12(-A\sin(x) + B\cos(x)) + 36(A\cos(x) + B\sin(x)) = 2\cos(x) - \sin(x)$

Simplify:

$-37A\cos(x) + 11B\sin(x) = 2\cos(x) - \sin(x)$

Now, equate the coefficients of the sine and cosine terms separately:

$-37A = 2 \quad \text{(coefficient of }\cos(x)\text{)}$

$11B = -1 \quad \text{(coefficient of }\sin(x)\text{)}$

Solving for $A$ and $B$:

$A = -\frac{2}{37}$

$B = -\frac{1}{11}$

Step 3: General Solution The general solution is the sum of the complementary and particular solutions:

$u(x) = u_c(x) + u_p(x) = (c_1 + c_2x)e^{-6x} - \frac{2}{37}\cos(x) - \frac{1}{11}\sin(x)$

where $c_1$ and $c_2$ are arbitrar