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Problem 7

For each of the initial-value problems use the method of successive approximations to find the first three members \(\phi_{1}, \phi_{2}, \phi_{3}\) of a sequence of functions that approaches the exact solution of the problem. $$ \frac{d y}{d x}=2 x+y^{3}, \quad y(0)=0 $$

Problem 7

Employ the method of isoclines to sketch the approximate integral curves of each of the differential equations $$ \frac{d y}{d x}=y^{3}-x^{2} $$

Problem 7

Obtain a power series solution in powers of \(x\) of each of the initial-value problems by (a) the Taylor series method and (b) the method of undetermined coefficients. $$ \frac{d y}{d x}=e^{x}+x \cos y, \quad y(0)=0 $$

Problem 8

For each of the initial-value problems use the method of successive approximations to find the first three members \(\phi_{1}, \phi_{2}, \phi_{3}\) of a sequence of functions that approaches the exact solution of the problem. $$ \frac{d y}{d x}=1+6 x y^{4}, \quad y(0)=0 $$

Problem 8

Employ the method of isoclines to sketch the approximate integral curves of each of the differential equations $$ \frac{d y}{d x}=\frac{3 x+2 y+x^{2}}{x+2 y} $$

Problem 9

Employ the method of isoclines to sketch the approximate integral curves of each of the differential equations $$ \frac{d y}{d x}=\frac{3 x+y+x^{3}}{5 x-y} $$

Problem 9

Obtain a power series solution in powers of \(x-1\) of each of the initial-value problems by (a) the Taylor series method and (b) the method of undetermined coefficients. $$ \frac{d y}{d x}=x^{3}+y^{2}, \quad y(1)=1 $$

Problem 10

Obtain a power series solution in powers of \(x-1\) of each of the initial-value problems by (a) the Taylor series method and (b) the method of undetermined coefficients. $$ \frac{d y}{d x}=x+y+y^{2}, \quad y(1)=1 $$

Problem 10

Employ the method of isoclines to sketch the approximate integral curves of each of the differential equations $$ \frac{d y}{d x}=\frac{\sin x+y}{x-y} $$

Problem 11

Employ the method of isoclines to sketch the approximate integral curves of each of the differential equations $$ \frac{d y}{d x}=\frac{\left(1-x^{2}\right) y-x}{y} $$

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