Question
If y1(x) is a solution of the differential equation dydx−f(x)y=0, then a solution of the differential equation dydx+f(x)y=r(x)
Answer: Option A
:
A
dydx−f(x).y=0dyy=f(x)dx
ln y=∫f(x)dx
y1(x)=e∫f(x)dxThen for given equation I.F = e∫f(x)dx
Hence Solution y.y1(x)=∫r(x).y1(x)dx
y=1y1(x)∫r(x).y1(x)dx
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:
A
dydx−f(x).y=0dyy=f(x)dx
ln y=∫f(x)dx
y1(x)=e∫f(x)dxThen for given equation I.F = e∫f(x)dx
Hence Solution y.y1(x)=∫r(x).y1(x)dx
y=1y1(x)∫r(x).y1(x)dx
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