Mister Exam

Differential equation 2y"+18y=6*tan(3t)

For Cauchy problem:

y() =
y'() =
y''() =
y'''() =
y''''() =

The graph:

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The solution

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    2                             
   d                              
2*---(y(t)) + 18*y(t) = 6*tan(3*t)
    2                             
  dt                              
$$18 y{\left(t \right)} + 2 \frac{d^{2}}{d t^{2}} y{\left(t \right)} = 6 \tan{\left(3 t \right)}$$
18*y + 2*y'' = 6*tan(3*t)
Detail solution
Divide both sides of the equation by the multiplier of the derivative of y'':
$$2$$
We get the equation:
$$9 y{\left(t \right)} + \frac{d^{2}}{d t^{2}} y{\left(t \right)} = 3 \tan{\left(3 t \right)}$$
This differential equation has the form:
y'' + p*y' + q*y = s,

where
$$p = 0$$
$$q = 9$$
$$s = - 3 \tan{\left(3 t \right)}$$
It is called linear inhomogeneous
second-order differential equation with constant coefficients.
The equation has an easy solution
We solve the corresponding homogeneous linear equation
y'' + p*y' + q*y = 0

First of all we should find the roots of the characteristic equation
$$q + \left(k^{2} + k p\right) = 0$$
In this case, the characteristic equation will be:
$$k^{2} + 9 = 0$$
Detailed solution of the equation
- this is a simple quadratic equation
The roots of this equation:
$$k_{1} = - 3 i$$
$$k_{2} = 3 i$$
As there are two roots of the characteristic equation,
and the roots are purely imaginary, then
solving the correspondent differential equation looks as follows:
$$y{\left(t \right)} = C_{1} \sin{\left(t \left|{k_{1}}\right| \right)} + C_{2} \cos{\left(t \left|{k_{2}}\right| \right)}$$
$$y{\left(t \right)} = C_{1} \sin{\left(3 t \right)} + C_{2} \cos{\left(3 t \right)}$$

We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y'' + p*y' + q*y = s

Use variation of parameters method
Suppose that C1 and C2 - it is functions of x

The general solution is:
$$y{\left(t \right)} = \operatorname{C_{1}}{\left(t \right)} \sin{\left(3 t \right)} + \operatorname{C_{2}}{\left(t \right)} \cos{\left(3 t \right)}$$
where C1(t) and C2(t)
by the method of variation of parameters, we find the solution from the system:
$$\operatorname{y_{1}}{\left(t \right)} \frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} + \operatorname{y_{2}}{\left(t \right)} \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} = 0$$
$$\frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} \frac{d}{d t} \operatorname{y_{1}}{\left(t \right)} + \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} \frac{d}{d t} \operatorname{y_{2}}{\left(t \right)} = f{\left(t \right)}$$
where
y1(t) and y2(t) - linearly independent particular solutions of Linear Ordinary Differential Equations,
y1(t) = sin(3*t) (C1=1, C2=0),
y2(t) = cos(3*t) (C1=0, C2=1).
The free term f = - s, or
$$f{\left(t \right)} = 3 \tan{\left(3 t \right)}$$
So, the system has the form:
$$\sin{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} + \cos{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} = 0$$
$$\frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} \frac{d}{d t} \sin{\left(3 t \right)} + \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} \frac{d}{d t} \cos{\left(3 t \right)} = 3 \tan{\left(3 t \right)}$$
or
$$\sin{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} + \cos{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} = 0$$
$$- 3 \sin{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} + 3 \cos{\left(3 t \right)} \frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} = 3 \tan{\left(3 t \right)}$$
Solve the system:
$$\frac{d}{d t} \operatorname{C_{1}}{\left(t \right)} = \sin{\left(3 t \right)}$$
$$\frac{d}{d t} \operatorname{C_{2}}{\left(t \right)} = - \sin{\left(3 t \right)} \tan{\left(3 t \right)}$$
- it is the simple differential equations, solve these equations
$$\operatorname{C_{1}}{\left(t \right)} = C_{3} + \int \sin{\left(3 t \right)}\, dt$$
$$\operatorname{C_{2}}{\left(t \right)} = C_{4} + \int \left(- \sin{\left(3 t \right)} \tan{\left(3 t \right)}\right)\, dt$$
or
$$\operatorname{C_{1}}{\left(t \right)} = C_{3} - \frac{\cos{\left(3 t \right)}}{3}$$
$$\operatorname{C_{2}}{\left(t \right)} = C_{4} + \frac{\log{\left(\sin{\left(3 t \right)} - 1 \right)}}{6} - \frac{\log{\left(\sin{\left(3 t \right)} + 1 \right)}}{6} + \frac{\sin{\left(3 t \right)}}{3}$$
Substitute found C1(t) and C2(t) to
$$y{\left(t \right)} = \operatorname{C_{1}}{\left(t \right)} \sin{\left(3 t \right)} + \operatorname{C_{2}}{\left(t \right)} \cos{\left(3 t \right)}$$
The final answer:
$$y{\left(t \right)} = C_{3} \sin{\left(3 t \right)} + C_{4} \cos{\left(3 t \right)} + \frac{\log{\left(\sin{\left(3 t \right)} - 1 \right)} \cos{\left(3 t \right)}}{6} - \frac{\log{\left(\sin{\left(3 t \right)} + 1 \right)} \cos{\left(3 t \right)}}{6}$$
where C3 and C4 is a constants
The answer [src]
                     /     log(1 + sin(3*t))   log(-1 + sin(3*t))\         
y(t) = C2*sin(3*t) + |C1 - ----------------- + ------------------|*cos(3*t)
                     \             6                   6         /         
$$y{\left(t \right)} = C_{2} \sin{\left(3 t \right)} + \left(C_{1} + \frac{\log{\left(\sin{\left(3 t \right)} - 1 \right)}}{6} - \frac{\log{\left(\sin{\left(3 t \right)} + 1 \right)}}{6}\right) \cos{\left(3 t \right)}$$
The classification
nth linear constant coeff variation of parameters
nth linear constant coeff variation of parameters Integral
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