© examsiri.com
Question : 2 of 48
Marks:
+1,
-0
Solution:
Given, differential equation
‌=‌ Let
x=X+h,y=Y+k, then
‌‌=‌| 2(X+h)+(Y+k)−3 |
| 2(Y+k)−(X+h)+3 |
‌=‌| 2X+Y+2h+k−3 |
| 2Y−X+2k−h+3 |
Let
2h+k−3=0 and
−h+2k+3=0 Solving these two equations, we get
h=‌ and
k=‌ So, Eq. (i) becomes,
‌=‌ Let
Y=vX ⇒‌=v+X‌ Substituting this into Eq. (ii), we get
⇒‌‌x‌‌=‌−v ‌⇒‌ ‌=‌ ⇒‌dv=‌ Integrating both sides, we get
∫‌dv=∫‌ Let
u=−2v2+2v+2 Then
du=(−4v+2)dv ∴‌‌∫‌=∫‌⇒‌‌ln|u|=ln|X|+c1, where
c1= constant
‌⇒‌‌‌‌ln‌|−2v2+2v+2|=ln|X|+c1 ‌⇒‌‌ln‌|−2v2+2v+2|=−2‌ln|X|+c2where
c2=−2c1‌⇒‌‌ln‌|X2(−2v2+2v+2)|=c2‌⇒‌‌X2(−2v2+2v+2)=c,‌ where ‌c=ec2‌⇒‌‌X2(‌+‌+2)=c‌⇒‌‌(−2Y2+2XY+2X2)=cSubstitute
X and
Y, we get
‌−2(y+‌)2+2(x−‌)(y+‌)+2(x−‌)2=c‌⇒−2(y2+‌y+‌)+2(xy+‌x−‌y−‌)+2(x2−‌x+‌)=c‌‌‌⇒‌‌−2y2−‌y−‌+2xy+‌x−‌y‌‌‌−‌+2x2−‌x+‌=c‌⇒−2y2+2xy+2x2−‌y−‌x+‌=c‌⇒2x2+2xy−2y2−6x−6y=c‌⇒x2+xy−y2−3x−3y=cThus, the general solution is
x2+xy−y2−3x−3y+c=0
© examsiri.com
Go to Question: