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Popular >

y^'=(4y+21)^{(1/2)}

Solution

y^{^{'}} = (4 y + 21)^{(\frac{1}{2})}

Solution

y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} + 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}, y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} - 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}

Solution steps

y^{'} (t) = (4 y + 21)^{(\frac{1}{2})}

Solve separable ODE:

y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} + 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}, y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} - 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}

y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} + 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}, y = \frac{- 21 + 2 c _{1}^{2} + 4 c _{1} t + 2 t ^{2} - 2 ( t ^{4} + 4 c _{1} t ^{3} + 6 c _{1}^{2} t ^{2} + 4 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{4}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=(4y+21)^{(1/2)} ?
The solution for y^'=(4y+21)^{(1/2)} is