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Popular >

y^'=(-4y+44)^{1/2}

Solution

y^{^{'}} = (- 4 y + 44)^{\frac{1}{2}}

Solution

y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} + ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}, y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} - ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}

Solution steps

y^{'} (t) = (- 4 y + 44)^{\frac{1}{2}}

Solve separable ODE:

y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} + ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}, y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} - ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}

y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} + ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}, y = \frac{88 - c _{1}^{2} - 4 c _{1} t - 4 t ^{2} - ( 16 t ^{4} + 32 c _{1} t ^{3} + 24 c _{1}^{2} t ^{2} + 8 c _{1}^{3} t + c _{1}^{4} ) ^{\frac{1}{2}}}{8}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=(-4y+44)^{1/2} ?
The solution for y^'=(-4y+44)^{1/2} is