What is the solution for 2xy^'+7y=7x^4,y(1)=4 ?

The solution for 2xy^'+7y=7x^4,y(1)=4 is y=(7x^4)/(15)+(53)/(15x^{7/2)}

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2xy^'+7y=7x^4,y(1)=4

2 x y^{^{'}} + 7 y = 7 x^{4}, y (1) = 4

y = \frac{7 x ^{4}}{15} + \frac{53}{15 x ^{\frac{7}{2}}}

Solution steps

2 x y^{'} (x) + 7 y = 7 x^{4}

Solve linear ODE:

y = \frac{7 x ^{4}}{15} + \frac{53}{15 x ^{\frac{7}{2}}}

y = \frac{7 x ^{4}}{15} + \frac{53}{15 x ^{\frac{7}{2}}}

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What is the solution for 2xy^'+7y=7x^4,y(1)=4 ?
The solution for 2xy^'+7y=7x^4,y(1)=4 is y=(7x^4)/(15)+(53)/(15x^{7/2)}