𝑥2𝑦=1+𝑐𝑥
differentiate both parts of the given equation respect to x:
(𝑥2𝑦)x′=(1+𝑐𝑥)x′(𝑥^2𝑦)_x'=(1+𝑐𝑥)_x'(x2y)x′=(1+cx)x′
we can get Differential Equation:
2xy+x2y′=c2xy+x^2y'=c2xy+x2y′=c
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments