1st order differential equation

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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