subzero0137
- 91
- 4
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?
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?