Q.
Let y(x) is the solution of the differential equation (x+2)dxdy−(x+1)y=2 . If y(0)=−1 , then the value of y(2) is equal to
2214
184
NTA AbhyasNTA Abhyas 2020Differential Equations
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Solution:
Given differential equation is dxdy−x+2x+1y=x+22
I.F. =e−∫x+2x+1dx=e−x(x+2)
So the solution of the equation is y(x+2)e−x=−2e−x+C
Since y(0)=1, we have C=4
So y(2)=e2−21