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This can be rewritten as dy 2 ± √ √ ₁ + ( ² ) ² , dx X so it is homogeneous. Putting y = xv gives +v=v± √²+v²₂ € SO = dv Ꮖ . dx Y how come ??

This can be rewritten as dy 2 ± √ √ ₁ + ( ² ) ² , dx X so it is homogeneous. Putting y = xv gives +v=v± √²+v²₂ € SO = dv Ꮖ . dx Y how come ??

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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First we solve the quadratic equation 2 dy (d) ² dx X SO dy Y to obtain the two possible values dy y + x² + y² dx x This can be rewritten as - dx X dv dx X- 2y so it is homogeneous. Putting y = xv gives +v=v± √√√1+v², D4 dy dx = which gives - x = 0 ± = √/₁ + (²) ²₁ dx dv [d²= = ± √ √ √1 + 0² X Thus In x + lnc = sinh v or v = sinh (ln(cx)). Hence Y eln(cx) e-ln(cx) X - 2 how come ?? 2cy = ± (c²x² − 1). = It is interesting to note that equation (1.14) also has the solution y tix, which cannot be obtained from the above solution. This is because it is a non-linear differential equation.
Transcribed Image Text:First we solve the quadratic equation 2 dy (d) ² dx X SO dy Y to obtain the two possible values dy y + x² + y² dx x This can be rewritten as - dx X dv dx X- 2y so it is homogeneous. Putting y = xv gives +v=v± √√√1+v², D4 dy dx = which gives - x = 0 ± = √/₁ + (²) ²₁ dx dv [d²= = ± √ √ √1 + 0² X Thus In x + lnc = sinh v or v = sinh (ln(cx)). Hence Y eln(cx) e-ln(cx) X - 2 how come ?? 2cy = ± (c²x² − 1). = It is interesting to note that equation (1.14) also has the solution y tix, which cannot be obtained from the above solution. This is because it is a non-linear differential equation.
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