Exercise:
Solve the following differential equations :
1) y″+y′−2y=(x−1)e−x
2) (1+x2)dydx=2x
Solution:
1) Consider the characteristic equation r2+r−2=0 for y″+y′−2y=0
⇒r2+r−2=(r+2)(r−1)=0 ⇒r=−2orr=1
So yH=c1e−2x+c2ex
yG=yH+yp
Take yp=(ax+b)e−x⇔y′p=ae−x−e−x(ax+b)
⇔y′′p=−ae−x−(−e−x(ax+b)+ae−x)
Now by substitution the particular solution we get the values for a and b
So −ae−x+(ax+b)e−x−ae−x+ae−x−(ax+b)e−x−2(ax+b)e−x=(x−1)e−x
⇒−ae−x−2(ax+b)e−x=(x−1)e−x⇒−a−2ax−2b=x−1
Hence −2a=1&−a−2b=−1⇔a=−12&b=34
Thus y(x)=c1e−2x+c2ex(−12x+34)e−x
2) We have (1+x2)dydx=2x ⇔(1+x2)dy=2xdx⇔2x1+x2dx=dy⇔∫2x1+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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