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Hi, everyone. Don’t know how to solve (x^2-1)y"+xy'-y=0
Which books are nice discussing about this kind of differential equations?
Many many thanks
A1. y\prime=\frac{4x^2}{y}+\frac{y}{x}
Ans: y=2x \sqrt{2x+3}
A2. y\prime=\frac{2y}{x}-\frac{x^4}{2y}
Ans: y=x^2 \sqrt{1-x}
B1. (1+x^2)y\prime\prime+3x y\prime =0
Ans: y=\frac{x}{\sqrt{1+x^2}}
B2. (x^2-1)y\prime\prime +x y\prime-y=0
Ans: y=x+ \sqrt{x^2-1}
B3. y\prime\prime +\frac{2}{x} y\prime +y=0
Ans: y=\frac{\cos{x}}{x}
Which books are nice discussing about this kind of differential equations?
Many many thanks
A1. y\prime=\frac{4x^2}{y}+\frac{y}{x}
Ans: y=2x \sqrt{2x+3}
A2. y\prime=\frac{2y}{x}-\frac{x^4}{2y}
Ans: y=x^2 \sqrt{1-x}
B1. (1+x^2)y\prime\prime+3x y\prime =0
Ans: y=\frac{x}{\sqrt{1+x^2}}
B2. (x^2-1)y\prime\prime +x y\prime-y=0
Ans: y=x+ \sqrt{x^2-1}
B3. y\prime\prime +\frac{2}{x} y\prime +y=0
Ans: y=\frac{\cos{x}}{x}
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