Take the homogeneous part and find the roots to the characteristic equation:
![y''+4y=0\implies r^2+4=0\implies r=\pm2i](https://tex.z-dn.net/?f=y%27%27%2B4y%3D0%5Cimplies%20r%5E2%2B4%3D0%5Cimplies%20r%3D%5Cpm2i)
This means the characteristic solution is
![y_c=C_1\cos2x+C_2\sin2x](https://tex.z-dn.net/?f=y_c%3DC_1%5Ccos2x%2BC_2%5Csin2x)
.
Since the characteristic solution already contains both functions on the RHS of the ODE, you could try finding a solution via the method of undetermined coefficients of the form
![y_p=ax\cos2x+bx\sin2x](https://tex.z-dn.net/?f=y_p%3Dax%5Ccos2x%2Bbx%5Csin2x)
. Finding the second derivative involves quite a few applications of the product rule, so I'll resort to a different method via variation of parameters.
With
![y_1=\cos2x](https://tex.z-dn.net/?f=y_1%3D%5Ccos2x)
and
![y_2=\sin2x](https://tex.z-dn.net/?f=y_2%3D%5Csin2x)
, you're looking for a particular solution of the form
![y_p=u_1y_1+u_2y_2](https://tex.z-dn.net/?f=y_p%3Du_1y_1%2Bu_2y_2)
. The functions
![u_i](https://tex.z-dn.net/?f=u_i)
satisfy
![u_1=\displaystyle-\int\frac{y_2(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx](https://tex.z-dn.net/?f=u_1%3D%5Cdisplaystyle-%5Cint%5Cfrac%7By_2%28%5Ccos2x%2B%5Csin2x%29%7D%7BW%28y_1%2Cy_2%29%7D%5C%2C%5Cmathrm%20dx)
![u_2=\displaystyle\int\frac{y_1(\cos2x+\sin2x)}{W(y_1,y_2)}\,\mathrm dx](https://tex.z-dn.net/?f=u_2%3D%5Cdisplaystyle%5Cint%5Cfrac%7By_1%28%5Ccos2x%2B%5Csin2x%29%7D%7BW%28y_1%2Cy_2%29%7D%5C%2C%5Cmathrm%20dx)
where
![W(y_1,y_2)](https://tex.z-dn.net/?f=W%28y_1%2Cy_2%29)
is the Wronskian determinant of the two characteristic solutions.
![W(\cos2x,\sin2x)=\begin{bmatrix}\cos2x&\sin2x\\-2\cos2x&2\sin2x\end{vmatrix}=2](https://tex.z-dn.net/?f=W%28%5Ccos2x%2C%5Csin2x%29%3D%5Cbegin%7Bbmatrix%7D%5Ccos2x%26%5Csin2x%5C%5C-2%5Ccos2x%262%5Csin2x%5Cend%7Bvmatrix%7D%3D2)
So you have
![u_1=\displaystyle-\frac12\int(\sin2x(\cos2x+\sin2x))\,\mathrm dx](https://tex.z-dn.net/?f=u_1%3D%5Cdisplaystyle-%5Cfrac12%5Cint%28%5Csin2x%28%5Ccos2x%2B%5Csin2x%29%29%5C%2C%5Cmathrm%20dx)
![u_1=-\dfrac x4+\dfrac18\cos^22x+\dfrac1{16}\sin4x](https://tex.z-dn.net/?f=u_1%3D-%5Cdfrac%20x4%2B%5Cdfrac18%5Ccos%5E22x%2B%5Cdfrac1%7B16%7D%5Csin4x)
![u_2=\displaystyle\frac12\int(\cos2x(\cos2x+\sin2x))\,\mathrm dx](https://tex.z-dn.net/?f=u_2%3D%5Cdisplaystyle%5Cfrac12%5Cint%28%5Ccos2x%28%5Ccos2x%2B%5Csin2x%29%29%5C%2C%5Cmathrm%20dx)
![u_2=\dfrac x4-\dfrac18\cos^22x+\dfrac1{16}\sin4x](https://tex.z-dn.net/?f=u_2%3D%5Cdfrac%20x4-%5Cdfrac18%5Ccos%5E22x%2B%5Cdfrac1%7B16%7D%5Csin4x)
So you end up with a solution
![u_1y_1+u_2y_2=\dfrac18\cos2x-\dfrac14x\cos2x+\dfrac14x\sin2x](https://tex.z-dn.net/?f=u_1y_1%2Bu_2y_2%3D%5Cdfrac18%5Ccos2x-%5Cdfrac14x%5Ccos2x%2B%5Cdfrac14x%5Csin2x)
but since
![\cos2x](https://tex.z-dn.net/?f=%5Ccos2x)
is already accounted for in the characteristic solution, the particular solution is then
![y_p=-\dfrac14x\cos2x+\dfrac14x\sin2x](https://tex.z-dn.net/?f=y_p%3D-%5Cdfrac14x%5Ccos2x%2B%5Cdfrac14x%5Csin2x)
so that the general solution is
He has 20 pencils.
The ratio in the problem is 4 pens to 5 pencils meaning for every 4 pens there are 5 pencils.
So take this into mathematics 16pens/4pens = 4
5pencils * 4 is equal to 20 pencils
the ratio still works and it is still 4 to 5 so the answer is 20...pencils.
Answer:
5ln3=ln(3^5)
Step-by-step explanation:
Given: 5ln(3)
Use rule: alog(b)=log(b^a), aln(b)=ln(b^a) (doesn't matter what the log base is)
Apply rule: ln(3^5)