1st order differential equation

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SUMMARY

The discussion focuses on solving the first order differential equation (y+x²y)dy/dx=3x+xy², with the initial condition y(1)=1. The solution process involves separating variables and integrating both sides, leading to the equation y²+3=x²+1+e^C. A correction is noted regarding the application of exponential rules, specifically a^(b+c)=a^b*a^c. The final solution derived is y=±x, after applying the boundary conditions.

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subzero0137
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Find the general solution of the first order differential equation (y+x^{2}y)\frac{dy}{dx}=3x+xy^{2}, with y(1)=1.



My attempt:
\frac{y}{3+y^{2}}dy=\frac{x}{1+x^{2}}dx ∴ \frac{1}{2}\int \frac{2y}{3+y^{2}}dy=\frac{1}{2}\int \frac{2x}{1+x^2}dx
=\frac{1}{2}ln|3+y^{2}|=\frac{1}{2}ln|1+x^{2}|+C ∴ y^{2}+3=x^{2}+1+e^{C} ∴ y=\pm\sqrt{x^{2}-2+e^{C}} Applying boundary conditions gives 1=\pm\sqrt{1^{2}-2+e^{C}} \Rightarrow e^{C}=2 Therefore y=\pm x.


Is this right?
 
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subzero0137 said:
y^{2}+3=x^{2}+1+e^{C}

Not quite right here. Pay particular attention to your exponential rules. Remember that

a^{b+c}=a^ba^c
 
Mentallic said:
Not quite right here. Pay particular attention to your exponential rules. Remember that

a^{b+c}=a^ba^c

Ohhh! Of course...silly me.
 
subzero0137 said:
Ohhh! Of course...silly me.

Don't worry, I answered a whole homework problem set on this topic with the exact same mistake in each and every question haha :biggrin:
 

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