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Linear Differential equation of first and second order ODE


Exercise:

Solve the following differential equations :

1) y+y2y=(x1)ex

2) (1+x2)dydx=2x

Solution:

1) Consider the characteristic equation r2+r2=0 for  y+y2y=0

r2+r2=(r+2)(r1)=0 r=2orr=1

So yH=c1e2x+c2ex

yG=yH+yp

Take yp=(ax+b)exyp=aexex(ax+b)

yp=aex(ex(ax+b)+aex)

Now by substitution the particular solution we get the values for a and b

So aex+(ax+b)exaex+aex(ax+b)ex2(ax+b)ex=(x1)ex

aex2(ax+b)ex=(x1)exa2ax2b=x1

Hence 2a=1&a2b=1a=12&b=34

Thus y(x)=c1e2x+c2ex(12x+34)ex

2) We have (1+x2)dydx=2x (1+x2)dy=2xdx2x1+x2dx=dy2x1+x2dx=dy

So y=2x1+x2dx=d(1+x2)1+x2=ln|1+x2|+c

Hence y(x)=ln|1+x2|+c

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