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  • Differential Equation Solutions

    Topics || Problems

    Obtain the general equation of \( \tan^2 y dy = \sin^3 x dx\)

    \( (\sec^2 y- 1)dy = \sin^2 x \sin x dx \)

    \( \int{(\sec^2 y- 1)dy} = \int{\sin^2 x \sin x dx}\)

    \( \tan y - y = \int{(1- \cos^2 x )( \sin x )}\)

    \( \tan y - y = \int{ \sin x dx }- \int{\cos^2 x \sin x dx} \)

    \( \tan y - y + c= -cos x + \frac{ \cos^3 x}{3} \)