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sergeinik [125]
3 years ago
7

(b) dy/dx = (x - y+ 1)^2

Mathematics
1 answer:
Elanso [62]3 years ago
3 0

Substitute v(x)=x-y(x)+1, so that

\dfrac{\mathrm dv}{\mathrm dx}=1-\dfrac{\mathrm dy}{\mathrm dx}

Then the resulting ODE in v(x) is separable, with

1-\dfrac{\mathrm dv}{\mathrm dx}=v^2\implies\dfrac{\mathrm dv}{1-v^2}=\mathrm dx

On the left, we can split into partial fractions:

\dfrac12\left(\dfrac1{1-v}+\dfrac1{1+v}\right)\mathrm dv=\mathrm dx

Integrating both sides gives

\dfrac{\ln|1-v|+\ln|1+v|}2=x+C

\dfrac12\ln|1-v^2|=x+C

1-v^2=e^{2x+C}

v=\pm\sqrt{1-Ce^{2x}}

Now solve for y(x):

x-y+1=\pm\sqrt{1-Ce^{2x}}

\boxed{y=x+1\pm\sqrt{1-Ce^{2x}}}

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