大一高等数学上册习题及答案.ppt

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1、2003级《高数》上试题解答Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.

2、0.Copyright2004-2011AsposePtyLtd.Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解先求函数y关于自变量x的导数Evaluationonly.CreatedwithAspose

3、.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解先求函数y关于自变量x的导数更准确：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解在隐函数方程的两边对x求导切线方程为解Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyr

4、ight2004-2011AsposePtyLtd.解：相应齐次特征方程为解得特征根为故齐次方程的通解为：非齐次方程有形如特解非齐次方程的通解为所求特解为Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解由于函数的连续性，分别计算函数值因此Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-201

5、1AsposePtyLtd.证Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.证只需设由介值定理知结论成立。Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5ClientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.解：Evaluationonly.CreatedwithAspose.Slidesfor.NET3.5C

6、lientProfile5.2.0.0.Copyright2004-2011AsposePtyLtd.