What is the solution for 3x^2y+x^3y^'=-7sec^2(x)tan(x) ?

The solution for 3x^2y+x^3y^'=-7sec^2(x)tan(x) is y=-(7tan^2(x))/(2x^3)+(c_{1})/(x^3)

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3x^2y+x^3y^'=-7sec^2(x)tan(x)

3 x^{2} y + x^{3} y^{^{'}} = - 7 sec^{2} (x) tan (x)

y = - \frac{7 tan ^{2} ( x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

Solution steps

3 x^{2} y + x^{3} y^{'} (x) = - 7 sec^{2} (x) tan (x)

Solve linear ODE:

y = - \frac{7 tan ^{2} ( x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

y = - \frac{7 tan ^{2} ( x )}{2 x ^{3}} + \frac{c _{1}}{x ^{3}}

f (x) = 2 x^{3} - 4 x + 6

\int_{0}^{4} 0 d x

y^{^{'}} = 4 + e^{y - 4 x + 1}

\frac{400}{s ^{2} + 12 s + 400}

f (x) = x^{2} + 4, (- 3, 13)

What is the solution for 3x^2y+x^3y^'=-7sec^2(x)tan(x) ?
The solution for 3x^2y+x^3y^'=-7sec^2(x)tan(x) is y=-(7tan^2(x))/(2x^3)+(c_{1})/(x^3)