Explanation
d2ydx2+3y=−2x.......(i)
This is second order non homogeneous differential equation.
Its solution is given as CF+PI
For CF part
d2ydx2+3y=0....(ii)y=c1emxdydx=c1memxd2ydx2=c1m2emx
Substituting in (ii), we get m2emx+3emx=0emx(m2+3)=0m2+3=0⇒m=√−3m=i√3y=c1ei√3x=c1(cos√3x+isin√3x)y=c1cos√3x+c2sin√3x
For PI part
Let y=cx+d
dydx=cd2ydx2=0
Substituting in (i)
0+3(cx+d)=−2x3cx+3d=−2x
Comparing both sides
c=−23,d=0
So solution for PI part is
y=−23x
General solution is CF+PI
c1cos√3x+c2sin√3x−23x
Given: ydydx=a−x
Integrating on both sides
∫ydx=∫adx−∫xdx
y22=ax−x22+c
x2+y2−2ax+c=0
(x−a)2+y2=r2
Hence family of circles with centre on x axis
A particle, initially at origin moves along x axis according to the rule dxdt=x+4. The time taken by the particle to traverse a distance of 96 units is
The equation of the curves, satisfying the differential equation d2ydx2(x2+1)=2xdydx passing through the point (0,1) and having the slope of tangent at x=0 as 6 is
e2x−3ydx+e2x−3ydy=0⇒e2xe3ydx+e2ye3xdy=0⇒e5xdx+e5ydye3ye3x=0⇒e5xdx+e5ydy=0
Integrating n both sides.
e5x5+e5y5=k⇒e5x+e5y=k
Given,
x2dydx.cos1x=−1
⇒dy=1cos1x.−1x2dx
Let us assume 1x=t.
Therefore, −1x2dx=dt.
⇒dy=1costdt
⇒dy=sectdt
Integrating both sides, we get,
⇒∫dy=∫sectdt
⇒y=loge|sect+tant|+C
⇒y=loge|sec1x+tan1x|+C
Now, given that y→−1asx→∞
⇒limx→∞y=−1
⇒limx→∞(loge|sec1x+tan1x|+C)=−1
⇒(loge|sec0+tan0|+C)=−1
⇒C=−1.
Therefore, the solution of given differential equation is:
y=loge|sec1x+tan1x|−1.
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