# [SOLVED]5.1.c trig IVP

#### karush

##### Well-known member
$\tiny{2.1.5.1.c}$ source
Change the second-order IVP into a system of equations
$\dfrac{d^2x}{dt^2}+\dfrac{dx}{dt}'+4x=\sin t \quad x(0)=4\quad x'(0)= -3$
ok I presume we can rewrite this as
$u''+u'+4u=\sin t$
Let $x_1=u$ and $x_2=u'$ then $x_1'=x_2$
substituting
$x_2'+x_2+4x=\sin t$
$\begin{array}{lllll} &let &x_1=u &and &x_2=u'\\ &then &x_1'=x_2 &and &x_2'=u'' \end{array}$
so
$\begin{array}{llll} x_1'=x_2\\ x_2'=-x_2-4x_1+\sin t \end{array}$

so far

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#### Country Boy

##### Well-known member
MHB Math Helper
Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.

#### karush

##### Well-known member
Since $x_2= x_1$ $x_2'= -5x_2+ sin(t)$. The associated homogeneous equation is $x_2'= \frac{dx}{dt}=-5x_2$ which we can write $\frac{dx_2}{x_2}= -5dt$. Integrating both sides, $ln(x_2)= -5t+ C$. Taking the exponential of both sides, $x_2= e^{-5t+ C}= e^Ce^{-5t}= C'e^{-5t}$.

Since the derivative of sin(x) is cos(x) and the derivative of cos(x) is -sin(x) we look for a solution to the entire equation of the form $x_2= A sin(x)+ B cos(x)$. Then $x_2'= A cos(x)- B sin(x)$ and the equation becomes
$A cos(x)- B sin(x)= -5(A sin(x)+ B cos(x))+ sin(x)$ so that $(A- B)cos(x)+ (5A- B)sin(x)= sin(x)$.

We must have A- B= 0 and 5A- B= 1.
mahalo that helped a lot....